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    NCERT Exemplar Class 12 Maths Solutions Chapter 12 Linear Programming

    Edited By Ravindra Pindel | Updated on Sep 15, 2022 05:19 PM IST | #CBSE Class 12th

    NCERT exemplar Class 12 Maths solutions chapter 12 Linear Programming - If you often wonder what the best solution to a set of problems would be, then linear programming would help you. It is the process of achieving the best possible outcome of a mathematical model. Explore the detailed concepts in NCERT exemplar Class 12 Maths chapter 12 solutions. Students looking for NCERT Exemplar Class 12 Maths solutions chapter 12 PDF download can use the ‘Save as webpage’ feature in the browser.
    Also, read - NCERT Class 12 Maths Solutions

    Question:1

    Determine the maximum value of Z = 11x + 7y subject to the constraints: 2x + y $ \leq $ 6, x $ \leq $ 2, x $ \geq $ 0, y $ \geq $ 0.

    Answer:

    Given that:
    Z = 11x+7y \\
    It is subject to constraints
    2x + y $ \leq $ 6, x $ \leq $ 2, x $ \geq $ 0, y $ \geq $ 0.
    Now let us convert given inequalities into equation.
    We obtain following equation
    \\ 2 x+y \leq 6 \\ \Rightarrow 2 x+y=6 \\ x \leq 2 \\ \Rightarrow x=2 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0
    The lines that represent 2x+y=6, then further meet the other axes respectively in order to get an answer. The points are to be joined to obtain the line 2x+y=6. It is then further clarified that the equation is satisfied. Then the region that contains the origin is then represented by the set of solutions of the inequation 2x+y$ \leq $ 6.\\
    The region represented by x$ \leq $ 2:\\
    The line which is parallel to the Y-axis then meets the X-axis which comes at X=2. Hence, it is clarified that (0,0) satisfies the inequation.
    After plotting the equation graphically, we get an answer:

    \\$The shaded region OBDE is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.\\ Corner Points are $\mathrm{O}(0,0), \mathrm{B}(0,6), \mathrm{D}(2,2)$ and $\mathrm{E}(2,0)$
    Coming to the conclusion, when we substitute the values in Z at the corner points, we get the following answer:
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=11 x+7 y \\ \hline O(0,0) & Z=11(0)+7(0)=0+0=0 \\ B(0,6) & Z=11(0)+7(6)=0+42=42 \rightarrow \max \\ D(2,2) & Z=11(2)+7(2)=22+14=36 \\ E(2,0) . & Z=11(2)+7(0)=22+0=22 \\ \hline \end{array}
    Therefore, the final answer is the maximum value of Z is 42 at the point (0,6).

    Question:2

    Maximise Z = 3x + 4y, subject to the constraints: x + y $ \leq $ 1, x $ \geq $ 0, y $ \geq $ 0. \\ \\

    Answer:

    Following is the answer
    Z = 3x + 4y
    It is subject to constraints
    x + y $ \leq $ 1, x $ \geq $ 0, y $ \geq $ 0. \\ \\
    Now let us convert the given inequalities into equation.
    We obtain the following equation
    \\x+y \leq 1 \\ \Rightarrow x+y=1 \\ x \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0
    The part represented by x+y$ \leq $ 1:
    One of the lines that is x+y=1 meets the axes (0,1) and (1,0) respectively. Then the lines are joined to obtain the line that is x+y=1. Therefore, it is clear that (0,0) the equation satisfies x+y$ \leq $ 1.
    The region that is represented by x$ \geq $ 0 \and\ y$ \geq $ 0 is first quadrant, and further satisfies these inequations. The graphic plotting is given below:

    The shaded region OBC shows the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
    Corner Points are $\mathrm{O}(0,0), \mathrm{B}(0,1)$ and $\mathrm{C}(1,0)$
    When we substitute the values in Z, we get the following answer
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=3 x+4 y \\ \hline O(0,0) & Z=3(0)+4(0)=0+0=0 \\ B(0,1) & Z=3(0)+4(1)=0+4=4 \rightarrow \max \\ C(1,0) & Z=3(1)+4(0)=3+0=3 \\ \hline \end{array}
    Hence, the maximum value of Z is 4 at the point (0,1).

    Question:3

    Maximize the function Z = 11x + 7y, subject to the constraints: x $ \leq $ 3, y $ \leq $ 2, x $ \geq $ 0, y $ \geq $ 0.\\

    Answer:

    It is subject to constraints
    x $ \leq $ 3, y $ \leq $ 2, x $ \geq $ 0, y $ \geq $ 0.\\
    Now let us convert the given inequalities into equation
    We obtain the following equation
    \\ x \leq 3 \\ \Rightarrow x=3 \\ y \leq 2 \\ \Rightarrow y=2 \\ y \geq 0 \\ \Rightarrow x=0 \\ y=0
    The region represented by x≤3:
    The line is parallel to the Y-axis and then meets the X-axis at the point x=3. Then it further gives a clarification that it satisfies the inequation in the problem that isx$ \leq $ 2. The region then represents the origin and the set of the inequation x$ \leq $ 3.\\
    The region that is represented by y$ \leq $ 2:\\
    The line that is parallel to the x-axis meets the y-axis. The part that contains the region represents the solution set of the other inequation y$ \leq $ 2.\\
    Therefore, the region that represents the x$ \geq $ 0 \ and\ y$ \geq $ 0 is first quadrant and satisfies the inequations. After plotting the graph we get


    The shaded region OBCD is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
    Corner Points are $\mathrm{O}(0,0), \mathrm{B}(0,2), \mathrm{C}(3,2)$ and $\mathrm{D}(3,0)$
    Now we will substitute these values in Z at each of these corner points, we get
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=11 x+7 y \\ \hline O(0,0) & Z=11(0)+7(0)=0+0=0 \\ B(0,2) & Z=11(0)+7(2)=0+14=14 \\ C(3,2) & Z=11(3)+7(2)=33+14=47 \rightarrow \max \\ D(3,0) & Z=11(3)+7(0)=33+0=33 \\ \hline \end{array}
    Therefore, the final answer is that the value of Z is 47 at the point of (3,2).

    Question:4

    Minimise Z = 13x -15y subject to the constraints: x + y $ \leq $ 7, 2x - 3y + 6 $ \geq $ 0, x $ \geq $ 0, y $ \geq $ 0.\\

    Answer:

    It is given that:
    Z = 13x -15y
    It is subject to constraints
    x + y $ \leq $ 7, 2x - 3y + 6 $ \geq $ 0, x $ \geq $ 0, y $ \geq $ 0.\\
    Now let us convert the given inequalities into equation
    We obtain the following equation
    \\ x+y \leq 7 \\ \Rightarrow x+y=7 \\ 2 x-3 y+6 \geq 0 \\ \Rightarrow 2 x-3 y+6=0 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ y=0
    The region which is represented by x+y$ \leq $ 7:\\
    The line in the sum meets the coordinate axes (7,0) and (0,7) respectively. If we join the lines, we will get the other line that is x + y =7. And then it is further clear that (0,0) satisfies the inequation. Then the origin represents the solution for the set of the inequation x+y$ \leq $ 7:\\
    The region represented by 2x - 3y + 6 $ \geq $ 0:\\
    The line 2x-3y+6=0 collides with the other axes to coordinate (-3,0) and (0,2) respectively. Then the lines are joined further to obtain the line 2x-3y+6=0. So, the part that contains the origin then represents the other solution set of the inequation 2x - 3y + 6 $ \geq $ 0. \\ \\
    Looking at the graph we get,

    The shaded region OBCD is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
    Corner Points are O(0,0), B(0,2), C(3,4) and D(7,0)
    Now we will substitute these values in Z at each of these corner points, we get
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=13 x-15 y \\ \hline O(0,0) & Z=13(0)-15(0)=0+0=0 \\ B(0,2) & Z=13(0)-15(2)=0-30=-30 \rightarrow \min \\ C(3,4) & Z=13(3)-15(4)=39-60=-21 \\ D(7,0) & Z=13(7)-15(0)=91-0=91 \end{array}
    So, the final answer of the question is the minimum value of Z is -30 at the point of (0,2).

    Question:5

    Determine the maximum value of Z = 3x + 4y if the feasible region (shaded) for a LPP is shown in Fig.12.7.

    Answer:

    From the question, it is given that
    Z= 3x+4y
    The figure that is given above, from that we can come to a constraint that
    \\ x + 2y $ \leq $ 76, 2x +y $ \leq $ 104, x $ \geq $ 0, y $ \geq $ 0\\
    Now let us convert the given inequalities into equation
    We obtain the following equation
    \\ x+2 y \leq 76 \\ \Rightarrow x+2 y=76 \\ 2 x+y \leq 104 \\ \Rightarrow 2 x+y=104 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0
    The region represented by x + 2y $ \leq $ 76:\\
    We can say that the line x + 2y=76 meets the coordinate axes (76,0) and (0,38) respectively. When we join the points to further get the required line we get the line x + 2y=76. Then, we can say that it is clear that (0,0) satisfies the inequation x + 2y $ \leq $ 76 . And then origin is represented by the solution set of inequationx + 2y $ \leq $ 76
    The region represented by 2x +y $ \leq $ 104:\\
    The line that has 2x +y=104 then meets the other coordinate axes (52,0) and (0,104) simultaneously. Then we need to join the points to get the result of 2x +y=104. The origin then represents the solution set further of the inequation 2x +y $ \leq $ 104:\\
    The first quadrant of the region represented is x$ \geq $ 0 and y$ \geq $ 0. \\ \\
    The graph of the equation is given below:

    The shaded region ODBA is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
    Corner Points are O(0,0), D(0,38), B(44,16) and A(52,0)
    Now we will substitute these values in Z at each of these corner points, we get
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=3 x+4 y \\ \hline O(0,0) & Z=3(0)+4(0)=0+0=0 \\ B(0,38) & Z=3(0)+4(38)=0+152=152 \\ C(44,16) & Z=3(44)+4(16)=132+64=196 \rightarrow \max \\ D(52,0) & Z=3(52)+4(0)=156+0=156 \\ \hline \end{array}
    Hence, the maximum value of Z is 196 at the point (44,16)

    Question:6

    Feasible region (shaded) for a LPP is shown in Fig. 12.8. Maximise Z = 5x + 7y.\\

    Answer:


    Following is the equation:
    Z=5x+7y
    The region that is shaded is OABD in the diagram that is being given. The maximum value of the corner point will occur at the feasible point.
    When we substitute these values in Z, we get the corner points, we get:
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } \mathrm{Z}=5 \mathrm{x}+7 \mathrm{y} \\ \hline \mathrm{O}(0,0) & \mathrm{Z}=5(0)+7(0)=0+0=0 \\ \mathrm{~A}(7,0) & \mathrm{Z}=5(7)+7(0)=35+0=35 \\ \mathrm{~B}(3,4) & \mathrm{Z}=5(3)+7(4)=15+28=43 \rightarrow \max \\ \mathrm{D}(0,2) & \mathrm{Z}=5(0)+7(2)=0+14=14 \\ \hline \end{array}
    Therefore, the final answer is the maximum value of Z is 43 at the point (3,4)

    Question:7

    The feasible region for a LPP is shown in Fig. 12.9. Find the minimum value of Z = 11x + 7y.

    Answer:

    The following is given that:
    Z = 11x + 7y.
    It is subject to constraints
    x+y \leq 5, x+3 y \geq 9, x \geq 0, y \geq 0
    Now let us convert the given inequalities into equation
    We obtain the following equation
    \\ x+y \leq 5 \\ \Rightarrow x+y=5 \\ x+3 y \geq 9 \\ \Rightarrow x+3 y=9 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0
    The region represented by x + y ≤ 5:
    The line that shows x+y=5, then meets the axes that coordinate (5,0) and (0,5) respectively. We need to join the points in order to obtain the line x+y=5. Hence, it is clarified that (0,0) then satisfies the inequation x + y $ \leq $ 5. Therefore, the region that has the origin represents the solution set of x + y $ \leq $ 5.
    The region that is represented by x +3y $ \geq $ 9:\\
    The line that is x+3y=9 meets the coordinate axes (9,0) and (0,3) respectively to get the final answer. When we join the points we get the line x+3y=9. Therefore, it is then clear that the region doesn’t contain the origin and represents the solution set of the inequation.
    The graph is further given below:

    The shaded region BEC is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
    Corner Points are B(0,3), E(0,5) and C(3,2)
    Now we will substitute these values in Z at each of these corner points, we get
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=11 x+7 y \\ \hline B(0,3) & Z=11(0)+7(3)=0+21=21 \rightarrow \min \\ E(0,5) & Z=11(0)+7(5)=0+35=35 \\ C(3,2) & Z=11(3)+7(2)=33+14=47 \\ \hline \end{array}
    Hence, the minimum value of Z is 21 at the point (0,3)

    Question:8

    Refer to Exercise 7 above. Find the maximum value of Z.

    Answer:

    It is given below that:
    Z=11x + 7y

    The figure that is given above, we can see the subject to constraints
    x + y $ \leq $ 5, x +3y $ \geq $ 9, x $ \geq $ 0, y $ \geq $ 0\\
    Now let us convert the given inequalities into equation
    We obtain the following equation
    \\ x+y \leq 5 \\ \Rightarrow x+y=5 \\ x+3 y \geq 9 \\ \Rightarrow x+3 y=9 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0
    The region which represents x + y $ \leq $ 5 is explained below:
    The line that is x+y=5, when meets the coordinate axes we get (9,0) and (0,3) respectively. After joining these points, we get the further line that is x+3y=9. The region does not represent the solution set.
    The region that represents the first quadrant is x$ \geq $ 0 and y$ \geq $ 0. \\
    The graph for the same is given below:

    The shaded region BEC is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
    Corner Points are B(0,3), E(0,5) and C(3,2)
    Now we will substitute these values in Z at each of these corner points, we get
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=11 x+7 y \\ \hline B(0,3) & Z=11(0)+7(3)=0+21=21 \\ E(0,5) & Z=11(0)+7(5)=0+35=35 \\ C(3,2) & Z=11(3)+7(2)=33+14=47 \rightarrow \max \\ \hline \end{array}
    \text { Hence, the maximum value of } Z \text { is } 47 \text { at the point }(3,2) \text { . }

    Question:9

    The feasible region for a LPP is shown in Fig. 12.10. Evaluate Z = 4x + y at each of the corner points of this region. Find the minimum value of Z, if it exists.

    Answer:

    It is subject to constraints
    x+2 y \geq 4, x+y \geq 3, x \geq 0, y \geq 0
    Now let us convert the given inequalities into equation
    We obtain the following equation
    \\ x+2 y \geq 4 \\ \Rightarrow x+2 y=4 \\ x+y \geq 3 \\ \Rightarrow x+y=3 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0\\ \Rightarrow y=0
    The region representing x + 2y $ \geq $ 4: \\
    The line that is x + 2y=4 meets the other coordinate axes (4,0) and (0,2) respectively. Once we join these points we get the result x + 2y=4. After joining the lines, it is then clarified that it does not satisfy the inequation x + 2y $ \geq $ 4: \\. So, then the region that does not contain the origin then represents the set of solutions.
    The region that is represented by x +y $ \geq $ 3:\\
    The other line that is x +y=3 meets the coordinate axes (3,0) and (0,3) respectively. When we join these points we obtain the line x +y=3. It is then cleared that it does not satisfy the inequation. Therefore, the region that does not contain the origin further represents the solution set.
    Region represented by $x \geq 0$\ and\ $y \geq 0$ is first quadrant, since every point in the first quadrant satisfies these inequations
    The graph of these equations is given.

    The shaded region ABC is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
    Corner Points are A(0,3), B(2,1), and C(4,0) .
    When we see the region, we can see that it is unbounded, and 3 is not the minimum value of Z.
    When we decide this issue, we can graph the inequality 4x + y < 3 and then further check that whether the resulting does has the open half no point in common, otherwise Z has no minimum value.
    The graph that is shown above, we can see that it is clear that there is no point in common with the region and therefore, Z has a minimum value of 3 at (0, 3).
    Henceforth, the minimum value of Z is 3 at the point (0, 3).

    Question:10

    In Fig. 12.11, the feasible region (shaded) for a LPP is shown. Determine the maximum and minimum value of Z = x + 2y

    Answer:

    It is given that:
    Z = x + 2y
    The figure that is given above, from that we can see that the shaded region is feasible and is bounded, so the maximum and the minimum point will occur at the corner point.
    \begin{aligned} &\text { Corner Points are } \mathrm{P}\left(\frac{3}{13}, \frac{24}{13}\right), \mathrm{Q}\left(\frac{3}{2}, \frac{15}{4}\right), \mathrm{R}\left(\frac{7}{2}, \frac{3}{4}\right) \text { and } \mathrm{S}\left(\frac{18}{7}, \frac{2}{7}\right)\\ &\text { Now we will substitute these values in } \mathrm{Z} \text { at each of these corner points, we get } \end{aligned}
    \begin{aligned} &\begin{array}{|l|l|} \hline \begin{array}{l} \text { Corner } \\ \text { Point } \end{array} & \text { Value of } \mathrm{Z}=\mathrm{x}+2 \mathrm{y} \\ \hline P\left(\frac{3}{13}, \frac{24}{13}\right) & 2=\left(\frac{3}{13}\right)+2\left(\frac{24}{13}\right)=\frac{3}{13}+\frac{48}{13}=\frac{51}{13} \\ Q\left(\frac{3}{2}, \frac{15}{4}\right) & 2=\left(\frac{3}{2}\right)+2\left(\frac{15}{4}\right)=\frac{3}{2}+\frac{15}{2}=\frac{18}{2}=9 \rightarrow \max \\ R\left(\frac{7}{2}, \frac{3}{4}\right) & 2=\left(\frac{7}{2}\right)+2\left(\frac{3}{4}\right)=\frac{7}{2}+\frac{3}{2}=\frac{10}{2}=5 \\ S\left(\frac{18}{7}, \frac{2}{7}\right) & 2=\left(\frac{18}{7}\right)+2\left(\frac{2}{7}\right)=\frac{18}{7}+\frac{4}{7}=\frac{22}{7}=3 \frac{1}{7} \rightarrow \min \\ \hline \end{array}\\ \end{aligned}
    \\$ Hence, the maximum value of Z is 9 at the point $\left(\frac{3}{2}, \frac{15}{4}\right) .$\\ And the minimum value of $Z$ is $3 \frac{1}{7}$ at the point $\left(\frac{18}{7}, \frac{2}{7}\right)$.

    Question:11

    A manufacturer of electronic circuits has a stock of 200 resistors, 120 transistors and 150 capacitors and is required to produce two types of circuits A and B. Type A requires 20 resistors, 10 transistors and 10 capacitors. Type B requires 10 resistors, 20 transistors and 30 capacitors. If the profit on type A circuit is Rs 50 and that on type B circuit is Rs 60, formulate this problem as a LPP so that the manufacturer can maximise his profit.

    Answer:

    We can assume that the manufacturer produces x units of type A circuits and y units of type B circuits. We have made the following table from the data that is given to us:
    \begin{array}{|l|l|l|l|} \hline & \text { Type A (X) } & \text { Type B (Y) } & \text { Maximum Stock } \\ \hline \text { Resistors } & 20 & 10 & 200 \\ \hline \text { Transistors } & 10 & 12 & 120 \\ \hline \text { Capacitors } & 10 & 30 & 150 \\ \hline \text { Profit } & \text { Rs. 50 } & \text { Rs. 60 } & \\ \hline \end{array}
    From the table, we can see that the profit becomes, Z=50x+60y.\\ \\
    The constraints that we got, i.e. the subject to the constraints,
    20x+10y$ \leq $ 200 [this is resistor constraint]
    When we divide it throughout by 10, we get:
    $ \Rightarrow $ 2x+y$ \leq $ 20$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(i)\\ \\
    And 10x+20y$ \leq $ 120 [this is transistor constraint]
    After that when we divide it through 10, we get
    $ \Rightarrow $ x+3y$ \leq $ 15$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(iii)\\ \\
    And x$ \geq $ 0, y$ \geq $ 0 [non-negative constraint]
    So, the maximum profit is Z=50x+60y,\\
    subject to 2x+y$ \leq $ 20,\\
    \\ x+2y$ \leq $ 12\\ x+3y$ \leq $ 15\\ x$ \geq $ 0, y$ \geq $ 0\\ \\

    Question:12

    A firm has to transport 1200 packages using large vans which can carry 200packages each and small vans which can take 80 packages each. The cost for engaging each large van is Rs 400 and each small van is Rs 200. Not more than Rs 3000 is to be spent on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimise cost.

    Answer:

    Let us consider that the firm has X number of large vans and Y number of small vans . We have made the following table:
    \begin{array}{|l|l|l|l|} \hline & \text { Large Van (X) } & \text { Small Van (Y) } & \begin{array}{l} \text { Maximum/Minim } \\ \text { um } \end{array} \\ \hline \text { Packages } & 200 & 80 & 1200 \\ \hline \text { cost } & \text { Rs.400 } & \text { Rs.200 } & \text { Rs. 3000 } \\ \hline \end{array}
    Looking at the table, we can see that the cost becomes Z=400x+200y.
    Now when we minimize the cost, i.e. minimize Z=400x+200y
    The subject to constraints are:
    200x+80y \geq1200
    When we divide it by 40, we get:
    $ \Rightarrow $ 5x+2y$ \geq $ 30$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(i)\\ \\
    $And 400 x+200 y \leq 3000$
    Now will divide throughout by 200, we get
    $\Rightarrow 2 x+y \leq 15$
    Also given the number of large vans cannot exceed the number of small vans
    $\Rightarrow x \leq y$
    And $x \geq 0, y \geq 0 [non-negative constraint]
    So, minimize cost we have to minimize Z =400 \mathrm{x}+200 \mathrm{y}$ subject to
    \\ $5 x+2 y \geq 30$\\ $$ 2 x+y \leq 15 $$\\ $x \leq y$\\ $x \geq 0, y \geq 0$

    Question:13

    A company manufactures two types of screws A and B. All the screws have to pass through a threading machine and a slotting machine. A box of Type A screws requires 2 minutes on the threading machine and 3 minutes on the slotting machine. A box of type B screws requires 8 minutes of threading on the threading machine and 2 minutes on the slotting machine. In a week, each machine is available for 60 hours.

    On selling these screws, the company gets a profit of Rs 100 per box on type A screws and Rs 170 per box on type B screws.

    Formulate this problem as a LPP given that the objective is to maximise profit.

    Answer:

    Let’s assume that the company manufactures X boxes of type A screws and Y boxes of type B screws. Look at the following table:
    \begin{array}{|l|l|l|l|} \hline & \begin{array}{l} \text { Type A Screws } \\ (\mathrm{x} \text { boxes }) \end{array} & \begin{array}{l} \text { Type B Screws } \\ (\mathrm{y} \text { boxes }) \end{array} & \begin{array}{l} \text { Maximum time } \\ \text { available on } \\ \text { each machine } \\ \text { in a week } \end{array} \\ \hline \begin{array}{l} \text { Time } \\ \text { required for } \\ \text { screws on } \\ \text { threading } \\ \text { machine } \end{array} & 2 & 8 & \begin{array}{l} 60 \text { hours } \\ =60 \times 60 \mathrm{~min} \\ =3600 \mathrm{~min} \end{array} \\ \hline \begin{array}{l} \text { Time } \\ \text { required for } \\ \text { screws on } \\ \text { slotting } \\ \text { machine } \end{array} & 3 & 2 & \begin{array}{l} 60 \text { hours } \\ =60 \times 60 \mathrm{~min} \\ =3600 \mathrm{~min} \end{array} \\ \hline \text { Profit } & \text { Rs } 100 &\text { Rs } 170 & \\ \hline \end{array}
    According to the table, we can see that profit becomes $\mathrm{Z}=100 \mathrm{x}+170 \mathrm{y}$
    Now, we have to maximize the profit, i.e., maximize $\mathrm{Z}=100 \mathrm{x}+170 \mathrm{y}$
    The constraints so obtained, i.e., subject to the constraints,
    $2 x+8 y \leq 3600[time constraints for threading machine]
    Now will divide throughout by 2, we get
    $\Rightarrow x+4 y \leq 1800$
    And $3 x+2 y \leq 3600$ [time constraints for slotting machine]
    $\Rightarrow 3 x+2 y \leq 3600 \ldots \ldots \ldots \ldots . .(i i)$
    And x \geq 0,y\geq 0 [non-negative constraint]
    So, to maximize profit we have to maximize $\mathrm{Z}=100 \mathrm{x}+170 \mathrm{y}$ subject to
    \\$x+4 y \leq 1800$\\ $3 x+2 y \leq 3600$\\ $x \geq 0, y \geq 0$\\

    Question:14

    A company manufactures two types of sweaters: type A and type B. It costs Rs 360 to make a type A sweater and Rs 120 to make a type B sweater. The company can make at most 300 sweaters and spend at most Rs 72000 a day. The number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100. The company makes a profit of Rs 200 for each sweater of type A and Rs 120 for every sweater of type B. Formulate this problem as a LPP to maximise the profit to the company.

    Answer:

    Let’s take an assumption that the company manufactures x number of type A sweaters and y number of type B. We have made a table for your understanding which is given below:
    \begin{array}{|l|l|l|l|} \hline & \begin{array}{l} \text { Type A Sweaters } \\ (\mathrm{X}) \end{array} & \begin{array}{l} \text { Type B Sweaters } \\ (\mathrm{Y}) \end{array} & \\ \hline \text { cost per day } & \text { Rs. } 360 & \text { Rs. } 120 & \text { Rs. } 72000 \\ \hline \begin{array}{l} \text { Number of } \\ \text { Sweaters } \end{array} & 1 & 1 & 300 \\ \hline \text { Profit } & \text { Rs.200 } & \text { Rs.120 } & \\ \hline \end{array}
    If we look at the table, we can see that the profit becomes Z=200x+120y
    If we have to maximize the profit, then maximize Z=200x+120y
    The constraints so obtained, i.e., subject to the constraints,
    The company spends at most Rs 72000 a day.
    $$ \therefore 360 x+120 y \leq 72000 $$
    Divide throughout by 120, we get
    $\Rightarrow 3 x+y \leq 600 \ldots$...(i)
    Also, company can make at most 300 sweaters.:
    $x+y \leq 300 \ldots$ (ii)
    Also, the number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100
    i.e., $y-x \leq 100$
    $\Rightarrow y \leq 100+x \ldots \ldots \ldots$ .....(iii)
    And x \geq 0,y\geq 0 [non-negative constraint]
    So, to maximize profit we have to maximizeZ=200x+120y subject to
    \\ 3 x+y \leq 600 \\$$ \begin{array}{l} x+y \leq 300 \\ y \leq 100+x \\ x \geq 0, y \geq 0 \end{array} $$

    Question:15

    A man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2per km on petrol. If he rides it at a faster speed of 80 km/hour, the petrol cost increases to Rs 3 per km. He has at most Rs 120 to spend on petrol and one hour’s time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.

    Answer:

    If we see the man rides his motorcycle for a distance of X km at a speed of 50km/hr then he will have to spend Rs.2 per km for petrol.
    If we assume that the man rides his motorcycle for a distance of Y km at a speed of 80km/hr, then he will need to spend Rs.3 per km on petrol.
    Assuming that he has to spend Rs.120 on petrol for a total distance so the constraint becomes,
    2x+3y\leq120.....(i)
    Now also given he has at most one hour's time for total distance to be covered, so the constraint becomes
    \frac{x}{50}+\frac{y}{80} \leq 1$ \{as distance =speed x time $\}$
    Now taking the LCM as 400, we get
    $\Rightarrow 8 x+5 y \leq 400$..........(ii)
    And $\mathrm{x} \geq 0, \mathrm{y} \geq 0$ [non-negative constraint]
    He want to find out the maximum distance travelled, here total distance, $\mathrm{Z}=\mathrm{x}+\mathrm{y}$
    Now, we have to maximize the distance, i.e., maximize $\mathrm{Z}=\mathrm{x}+\mathrm{y}$
    So, to maximize distance we have to maximize, $\mathrm{Z}=\mathrm{x}+\mathrm{y}$, subject to
    \\$2 x+3 y \leq 120$ \\$8 x+5 y \leq 400$ \\$x \geq 0, y \geq 0$

    Question:16

    Refer to Exercise 11. How many circuits of Type A and of Type B, should be produced by the manufacturer so as to maximise his profit? Determine the maximum profit.

    Answer:

    If we refer to Exercise 11, we get the following information.
    The manufacturer produces X units of type A circuits and y units of type B circuits. We make the following table from the given data:
    \begin{array}{|l|l|l|l|} \hline & \text { Type A (X) } & \text { Type B (Y) } & \text { Maximum Stock } \\ \hline \text { Resistors } & 20 & 10 & 200 \\ \hline \text { Transistors } & 10 & 12 & 120 \\ \hline \text { Capacitors } & 10 & 30 & 150 \\ \hline \text { Profit } & \text { Rs. 50 } & \text { Rs. 60 } & \\ \hline \end{array}
    Looking at the table, we can see that profit becomes Z=50x+60y.
    When we maximize the profit, i.e. maximize Z=50x+60y.
    If we see at the constraints that we have obtained, then the subject to constraints,
    20x+10y$ \leq $ 200 [this is resistor constraint]
    Dividing it throughout by 10, we get the answer as:
    2x+y$ \leq $ 20$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(i)\\ \\
    And 10x+20y$ \leq $ 120 [this is transistor constraint]
    Dividing it by 10 throughout we get
    x+2y$ \leq $ 12$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(ii)\\ \\
    And 10x+30y$ \leq $ 150 [this is capacitor constraint]
    Then divide it by 10, we get
    x+3y$ \leq $ 15$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(iii)\\ \\
    And x$ \geq $ 0, y$ \geq $ 0 [non-negative constraint]
    So, when we look at the maximize profit it is Z=50x+60y, subject to
    2 x+y$ \leq $ 20\\ \\
    \\ x+2y$ \leq $ 12\\ \\ x+3y$ \leq $ 15\\ \\ x$ \geq $ 0, y$ \geq $ 0\\ \\
    When we convert it to equation, we get the following equation
    \\ 2x+y$ \leq $ 20\\ \\ 2x+y= 20\\ \\ x+2y$ \leq $ 12\\ \\ x+2y= 12\\ \\ x+3y$ \leq $ 15\\ \\ x+3y= 15\\ \\ x $ \geq $ 0\\ \\ x=0\\ \\ y $ \geq $ 0\\ \\ y=0\\ \\
    The region that is represented by 2x+y\leq 20:
    When the line 2x+y=20 meets the coordinate axes (10,0) and (0,20) respectively. Once after the lines are joined, we obtain the line 2x+y=20. It is clear that (0,0) satisfies the inequation 2x+y\leq 20. So the region then represents the set of solutions.
    The region represented by x+2y\leq 12:
    Once the line x+2y=12 meets the coordinate axes (12,0) and (0,6) respectively, then the points are joined to get the final result that is x+2y=12. It is then clear that it satisfies the inequation and the region does not contain the origin.
    The region represented by x+3y\leq15:
    The line x+3y=15 then meets the coordinate axes (15,0) and (0,5) respectively. Once we join these points we obtain the line x+3y=15. It is clear that (0,0) satisfies the inequation x+3y≤ 15. So the region that contains the origin represents the solution set of the inequation x+3y\leq15:.
    Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations
    The graph of the same is given below:

    The shaded region OABCD is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
    Corner Points are O(0,0), A(0,5) , B(6,3), C(9.3,1.3) and D(10,0)
    Now we will substitute these values in Z at each of these corner points, we get
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=50 x+60 y \\ \hline 0(0,0) & Z=50(0)+60(0)=0+0=0 \\ A(0,5) & Z=50(0)+60(5)=0+300=300 \\ B(6,3) & Z=50(6)+60(3)-300+180=480 \\ C(9.3,1.3) & Z=50(9.3)+60(1.3)=465+78=543 \rightarrow \max \\ D(10,0) & Z=50(10)+60(0)=500+0=500 \\ \hline \end{array}
    So from the above table the maximum value of Z is at point (9.3,1.3), but as the manufacturer is required to produce two type of circuits, so the parts of resistors, transistors and capacitors cannot be decimals. So we will consider the next maximum number.
    Hence, the maximum value of Z is 480 at the point (6,3) i.e., the manufacturer should produce 6 circuits of type A and 3 circuits of type B so as to maximize his profit.

    Question:17

    Refer to Exercise 12. What will be the minimum cost?

    Answer:

    We get the following information referring to the exercise 12:
    Let us consider that the firm has X number of large vans and Y number of small vans . We have made the following table:
    \begin{aligned} &\text { - }\\ &\begin{array}{|l|l|l|l|} \hline & \text { Large Van (X) } & \text { Small Van (Y) } & \begin{array}{l} \text { Maximum/Minim } \\ \text { um } \end{array} \\ \hline \text { Packages } & 200 & 80 & 1200 \\ \hline \text { cost } & \text { Rs.400 } & \text { Rs.200 } & \text { Rs. 3000 } \\ \hline \end{array} \end{aligned}
    According to the table, the cost becomes Z=400x+200y.
    When we have to minimize the cost, i.e., minimize Z=400x+200y.
    The subject to the constraints are as follows:
    200x+80y$ \geq $ 1200\\
    Then divide it by 40, we get
    5x+2y$ \geq $ 30$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(i)\\
    $ And 400x+200y$ \leq $ 3000\\
    Then divide it by 200, we get
    2x+y$ \leq $ 15$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(ii)\\
    The number of the vans that are large exceed the number of small vans
    \\ x$ \leq $ y$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(iii)\\ \\ $ And x$ \geq $ 0, y$ \geq $ 0 [non-negative constraint]\\
    So, when we minimize cost we have to minimize, Z=400x+200y, subject to
    \\ 5x+2y$ \geq $ 30\\ \\ 2x+y$ \leq $ 15\\ \\ x$ \leq $ y\\ \\ x$ \geq $ 0, y$ \geq $ 0\\ \\
    When we convert the inequalities into equation, we get the following equation
    \\ \\ 5x+2y$ \geq $ 30\\ \\ 5x+2y=30\\ \\ 2x+y$ \leq $ 15\\ \\ 2x+y=15\\ \\ x$ \leq $ y\\ \\ x=y\\ \\ x $ \geq $ 0\\ \\ x=0\\ \\ y $ \geq $ 0\\ \\ y=0\\ \\
    We can see the region represented by 5x+2y\geq 30:
    The line that is 5x+2y=30 when it meets the coordinate axes (6,0) and (0,15) respectively we get the desired answer. We will then join these points to obtain the line 5x+2y=30. It is clear that (0,0) does not satisfy the inequation 5x+2y\geq 30. So the region that does not contain the origin represents the solution set of the inequation 5x+2y\geq 30.
    The region that represents 2x+y\leq15:
    We can see that the line 2x+y=15 meets the coordinate axes (7.5,0) and (0,15) respectively. We will have to join these points to obtain the line 2x+y=15. It is clear that (0,0) satisfies the inequation 2x+y\leq15. So the region that contains the origin represents the solution set of the inequation 2x+y\leq15.
    The region represented by x≤y:
    Then when the line x=y is a line that passes through the origin and doesn’t touch any coordinate axes at any other point except (0,0). When we join the points we get line x=y.
    Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations

    The shaded region ABC represents the feasible region is bounded, and minimum value will occur at a corner point of the feasible region.
    \\ $ Corner Points are $ \mathrm{A}\left(\frac{30}{7}, \frac{30}{7}\right), \mathrm{B}(0,15) and \mathrm{C}(5,5)
    Now we will substitute these values in Z at each of these corner points, we get
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=400 \times+200 \mathrm{y} \\ \hline A\left(\frac{30}{7}, \frac{30}{7}\right) & 2=400\left(\frac{30}{7}\right)+200\left(\frac{30}{7}\right) \\ & =\frac{12000}{7}+\frac{6000}{7}=\frac{18000}{7} \\ B(0,15) & =2571.43 \rightarrow \min \\ C(5,5) & Z=400(0)+200(15)=0+3000=3000 \\ & Z=400(5)+200(5)=2000+1000=3000 \\ \hline \end{array}
    So from the above table the minimum value of Z is at point \left(\frac{30}{7}, \frac{30}{7}\right)$
    Hence, the minimum cost of the firm is Rs. 2571.43

    Question:18

    Refer to Exercise 13. Solve the linear programming problem and determine the maximum profit to the manufacturer.

    Answer:

    Refer to the Exercise 13, we get the following data:
    Let us assume that the company manufactures x boxes of type A screws and y boxes of type B screws. We make the following table from the given data:


    Type A screws (x boxes)

    Type B screws (y boxes)

    Max time available on each machine in a week

    Time required for screws on threading machine

    2

    8

    60 hrs = 60*60min = 3600min

    Time required for screws on slotting machine

    3

    3

    60 hrs = 60*60min = 3600min

    Profit

    Rs 100

    Rs170


    When we look at the table, the profit becomes, Z=100x+170y
    Thus according to the table, the profit becomes, Z=100x+170y
    The constraints that we have obtained that is subject to constraints:
    2x+8y$ \leq $ 3600 [time constraints for threading machine]
    Divide it throughout by 2, we get
    x+4y$ \leq $ 1800$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(i)\\
    And 3x+2y$ \leq $ 3600 [time constraints for slotting machine]
    3x+2y$ \leq $ 3600…………..(ii)
    And x≥0, y≥0 [non-negative constraint]
    So, to maximize profit we have to maximize, Z=100x+170y, subject to
    \\ \\ x+4y$ \leq $ 1800\\ \\ 3x+2y$ \leq $ 3600\\ \\ x$ \geq $ 0, y$ \geq $ 0\\
    Now let us convert the given inequalities into equation.
    We obtain the following equation
    \\ x+4y$ \leq $ 1800\\ \\ x+4y=1800\\ \\ 3x+2y$ \leq $ 3600\\ \\ 3x+2y=3600\\ \\ x $ \geq $ 0\\ \\ x=0\\ \\ y $ \geq $ 0\\ \\ y=0\\ \\
    The region that represents x+4y≤ 1800:
    We can say that the line x+4y=1800 meets the coordinate axes (1800,0) and (0,450) respectively. When we join these points, we get the desired line that is x+4y=1800. We can say that it is clear that it satisfies the inequation and therefore the region that contains the origin further represents the set of solutions of the inequation.
    The region represented by 3x+2y\leq3600:
    The line 3x+2y=3600 further meets the axes (1200,0) and (0,1800) respectively. After joining these points, we get the result that is 3x+2y\leq3600. It is then clear that it satisfies the inequation 3x+2y\leq3600. So the region that contains the origin represents the solution set of the inequation 3x+2y\leq3600.
    The graph is given below:


    The shaded region OBCD is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
    Corner Points are O(0,0), B(0,450), C(1080,180) and D(1200,0)
    Now we will substitute these values in Z at each of these corner points, we get
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=100 x+170 y \\ \hline O(0,0) & Z=100(0)+170(0)=0+0=0 \\ B(0,450) & Z=100(0)+170(450)=0+76500=76500 \\ C(1080,180) & Z=100(1080)+170(180)=108000+30600 \\ & Z=138600 \rightarrow \max \\ D(1200,0) & Z=100(1200)+170(0)=120000+0=120000 \\ \hline \end{array}
    So from the above table the maximum value of Z is at point (1080,180) .
    Therefore, the final answer is maximum profit to the manufacturer is Rs.
    1,38,600

    Question:19

    Refer to Exercise 14. How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit?

    Answer:

    When we refer to the exercise, we get the following information:
    Let’s take an assumption that the company manufactures x number of type A sweaters and y number of type B. We have made a table for your understanding which is given below:
    \begin{array}{|l|l|l|l|} \hline & \begin{array}{l} \text { Type A Sweaters } \\ (\mathrm{X}) \end{array} & \begin{array}{l} \text { Type B Sweaters } \\ (\mathrm{Y}) \end{array} & \\ \hline \text { cost per day } & \text { Rs. } 360 & \text { Rs. } 120 & \text { Rs. } 72000 \\ \hline \begin{array}{l} \text { Number of } \\ \text { Sweaters } \end{array} & 1 & 1 & 300 \\ \hline \text { Profit } & \text { Rs.200 } & \text { Rs.120 } & \\ \hline \end{array}
    Thus according to the table, the profit becomes, Z=200x+120y
    If we have to maximize the profit, i.e., maximize Z=200x+120y
    The subject to constraints is:
    The company spends at most Rs 72000 a day
    . $\therefore 360 x+120 y \leq 72000$
    Divide throughout by 120, we get
    $\Rightarrow 3 x+y \leq 600 \ldots$...(i)
    Also, company can make at most 300 sweaters.:$x+y \leq 300 \ldots$.. (ii)
    Also, the number of sweaters of type $B$ cannot exceed the number of sweaters of type A by more than 100
    i.e., $y-x \leq 100$
    $y \leq 100+x \ldots \ldots \ldots$(iii)
    And $x \geq 0, y \geq 0$ [non-negative constraint]
    \begin{aligned} &\text { So, to maximize profit we have to maximize, } Z=200 x+120 y, \text { subject to }\\ &3 x+y \leq 600\\ &x+y \leq 300\\ &y \leq 100+x\\ &x \geq 0, y \geq 0 \end{aligned}
    So, when we convert the inequalities into the equation we get the following answer:
    \\ \mathrm{3x}+\mathrm{y} \leq 600 \\ \Rightarrow 3 \mathrm{x}+\mathrm{y}=600 \\ \mathrm{x}+\mathrm{y} \leq 300 \\ \Rightarrow \mathrm{x}+\mathrm{y}=300 \\ \mathrm{y} \leq 100+\mathrm{x} \\ \Rightarrow \mathrm{y}=100+\mathrm{x} \\ \mathrm{x} \geq 0 \\ \\ \Rightarrow \mathrm{x}=0
    \begin{aligned} &y \geq 0\\ &\Rightarrow y=0\\ &\text { The region represented by } 3 \mathrm{x}+\mathrm{y} \leq 600 \text { : } \end{aligned}
    We can say that the line 3x+y=600 meets the coordinate axes (200,0) and (0,600) respectively. Once the points are joined, we get the answer that is 3x+y=600. It is justified that 3x+y≤ 600. So the region that contain the origin represents the solution set of the inequation 3x+y≤ 600
    The region represented by x+y≤ 300:
    The line when it meets the coordinate axes respectively that is y= 100+x meets (-100,0) and (0,100) respectively and when the lines are joined we get the line y= 100+x. Then it is clarified that it satisfies the inequation. Therefore, the region contains the origin that represents the solution of sets of inequation.
    The graph of the same is given:

    The shaded region OBCDE is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
    Corner Points are O(0,0), B(0,100), C(100,200) , D(150,150) and E(200,0)
    Now we will substitute these values in Z at each of these corner points, we get
    \begin{aligned} &\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=200 x+120 y \\ \hline O(0,0) & Z=200(0)+120(0)=0+0=0 \\ B(0,100) & Z=200(0)+120(100)=0+12000=12000 \\ C(100,200) & Z=200(100)+120(200)=20000+24000=44000 \\ D(150,150) & Z=200(150)+120(150)=30000+18000=48000 \rightarrow \\ E(200,0) & Z=200(200)+120(0)=40000+0=40000 \\ \hline \end{array}\\ &\text { So from the above table the maximum value of } Z \text { is at point }(150,150) \text { . } \end{aligned}
    Hence, the final answer is the maximum profit to the manufacturer is Rs. 48,000, for making 150 sweaters each of type A and type B.

    Question:20

    Refer to Exercise 15. Determine the maximum distance that the man can travel.

    Answer:

    When we refer to the Exercise, we get the following information:
    If the man rides his motorcycle for a distance of x km at a speed of 50km/hr then he has to spend Rs. 2/km on petrol.
    And if he rides his motorcycle for a distance of y km at a speed of 80 km/hr then he has to spend Rs. 3/km on petrol.
    He has Rs.120 to spend on petrol for covering the total distance so the constraint becomes
    2x+3y≤120…………(i)
    Given that he has only one hour’s time to cover the total distance, then the constraint becomes
    \frac{x}{50}+\frac{y}{80} \leq 1 {as distance speed x time}
    Now taking the LCM as 400, we get
    $\Rightarrow 8 x+5 y \leq 400 \ldots \ldots \ldots \ldots$(ii)
    And $\mathrm{x} \geq 0, \mathrm{y} \geq 0$ [non-negative constraint]
    He want to find out the maximum distance travelled, here total distance, $\mathrm{Z}=\mathrm{x}+\mathrm{y}$
    Now, we have to maximize the distance, i.e., maximize $\mathrm{Z}=\mathrm{x}+\mathrm{y}$
    So, to maximize distance we have to maximize, $\mathrm{Z}=\mathrm{x}+\mathrm{y}$,$ subject to
    \\$2 x+3 y \leq 120$ \\$8 x+5 y \leq 400$ \\$x \geq 0, y=0$
    When we convert it into equation we get
    \\ 2x+3y$ \leq $ 120$ \Rightarrow $ 2x+3y=120\\ \\ 8x+5y$ \leq $ 400 $ \Rightarrow $ 8x+5y=400\\ \\ x $ \geq $ 0 $ \Rightarrow $ x=0\\ \\ y $ \geq $ 0 $ \Rightarrow $ y=0\\ \\
    The region that is represented by 2x+3y≤120:
    The line that is 2x+3y=120 further meets the coordinate axes (60,0) and (0,40) respectively and then when we will join these points we obtain the line 2x+3y=120. It is clear that (0,0) justifies the inequation 2x+3y≤120. So the region that contain the origin represents the solution set of the inequation 2x+3y≤120
    The region represented by 8x+5y≤400:
    We can see that the line 8x+5y=400 meets the coordinate axes (50,0) and (0,80) respectively. When we join these points we obtain the line 8x+5y=400. Therefore, it is clear that (0,0) satisfies the inequation 8x+5y≤400. So the region that contain the origin represents the solution set of the inequation 8x+5y≤400
    The first quadrant of the region represented is x≥0 and y≥0. The graph is as follows:

    The shaded region OBCD is the feasible region is bounded, and maximum value will occur at a corner point of the feasible region.
    Corner Points are \mathrm{O}(0,0), \mathrm{B}(0,40), \mathrm{C}\left(\frac{300}{7}, \frac{90}{7}\right)$ and $\mathrm{D}(50,0)$
    When we substitute the values in Z, we get the following answer:
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=x+y \\ \hline O(0,0) & Z=0+0=0 \\ B(0,40) & Z=0+40=40 \\ C\left(\frac{300}{7}, \frac{00}{7}\right) & z=\frac{300}{7}+\frac{80}{7}=\frac{380}{7}=54 \frac{2}{7} \rightarrow \max \\ D(50,0) & Z=50+0=50 \\ \hline \end{array}
    So from the above table the maximum value of Z is at point \left(\frac{300}{7}, \frac{80}{7}\right)$
    Hence, the maximum distance the man can travel is 54 \frac{2}{7} \mathrm{~km}$ or 54.3 km

    Question:21

    Maximise Z = x + y subject to \\ x + 4y $ \leq $ 8, 2x + 3y $ \leq $ 12, 3x + y $ \leq $ 9, x $ \geq $ 0, y $ \geq $ 0.\\

    Answer:

    It is given that:
    Z = x + y
    And it is also subject to constraints that is given below:
    \\ x + 4y $ \leq $ 8\\2x + 3y $ \leq $ 12\\3x + y $ \leq $ 9\\ x $ \geq $ 0\\ y $ \geq $ 0.\\
    We have to maximize Z, we are subject to the constraints above.
    We need to convert the inequalities into equation to get the following equation:
    \\ \\ x + 4y $ \leq $ 8\\ \\ $ \Rightarrow $ x + 4y = 8\\ \\ 2x + 3y $ \leq $ 12\\ \\ $ \Rightarrow $ 2x + 3y = 12\\ \\ 3x + y $ \leq $ 9\\ \\ $ \Rightarrow $ 3x + y = 9\\ \\ x $ \geq $ 0\\ \\ $ \Rightarrow $ x=0\\ \\ y $ \geq $ 0\\ \\ $ \Rightarrow $ y=0\\
    The region that represents x + 4y $ \leq $ 8 is explained below:
    The line x + 4y = 8 meets the coordinate axes (8,0) and (0,2) respectively. We will join these points to obtain the line x + 4y = 8. It is clear that (0,0) satisfies the inequation x + 4y $ \leq $ 8. So, the region containing the origin represents the solution set of the inequation x + 4y $ \leq $ 8
    The region that represents 2x + 3y \leq 12:
    The line that is 2x + 3y = 12 then meets the coordinate axes respectively to get the answer. When we join the points we obtain the line 2x + 3y = 12. It is clear that (0,0) satisfies the inequation 2x + 3y \leq 12. So, the region containing the origin represents the solution set of the inequation2x + 3y \leq 12.
    The region that represents 3x + y \leq 9:
    The line meets the coordinate axes that is 3x + y = 9 meets (3,0) and (0,9) respectively. After joining the lines, we get 3x + y = 9 and then it is clear that (0,0) satisfies the inequation. The region that contains the origin is represented by the solution set of 3x + y \leq 9.
    The graph for the same is given below and also the final answer:

    \\$Feasible region is ABCD\\ Value of $Z$ at corner points $A, B, C$ and $D-$
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=x+y \\ \hline \text { A }(2,0) & z=2+0=2 \\ \hline B(2.54,1.36) & Z=2.54+1.36=3.90 \rightarrow \max \\ \hline \text { C }(3,0) & z=3+0=3 \\ \hline \text { D }(0,0) & Z=0+0=0 \\ \hline \end{array}
    \text { So, value of } Z \text { is maximum at } B(2.54,1.36), \text { the maximum value is } 3.90 .

    Question:22

    A manufacturer produces two Models of bikes - Model X and Model Y. Model X takes 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hours available per week. Handling and Marketing costs are Rs 2000 and Rs 1000 per unit for Models X and Y respectively. The total funds available for these purposes are Rs 80,000 per week. Profits per unit for Models X and Y are Rs 1000 and Rs 500, respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.

    Answer:

    Let us say that the number of bikes per week of model X and Y are x and y respectively.
    Assuming that model X takes 6 man-hours.
    So, time taken by x bikes of model X = 6x hours.
    Assuming that model Y takes 10 man-hours.
    So, time taken by y bikes of model X = 10y hours.
    So, the total man-hour that is available per week = 450
    So, 6x + 10y ≤ 450
    3x + 5y ≤ 225
    The handling and the marketing cost of model X and Y is Rs 2000 and Rs 1000 per unit respectively.
    So, total handling and marketing cost of x units of model X and y units of model y is 2000x + 1000y
    The maximum amount that is available for handling and marketing per week is Rs 80000.
    So, 2000x + 1000y \leq 80000
    \Rightarrow 2x + y ? 80
    Profits per unit for Models X and Y are Rs 1000 and Rs 500, Respectively.
    Let total profit = Z
    So, Z = 1000x + 500y
    Also, as units will be positive numbers so x, y ≥ 0
    So, we have,
    Z = 1000x + 500y
    With constraints,
    \\ \\ 3x + 5y $ \leq $ 225\\ \\ 2x + y $ \leq $ 80\\ \\ x, y $ \geq $ 0\\ \\
    In order to maximize Z, that is subject to constraints.
    We need to convert it into equation:
    \\ 3x + 5y $ \leq $ 225\\ \\ $ \Rightarrow $ 3x + 5y = 225\\ \\ 2x + y $ \leq $ 80\\ \\ $ \Rightarrow $ 2x + y = 80\\ \\ x $ \geq $ 0\\ \\ $ \Rightarrow $ x=0\\ \\ y $ \geq $ 0\\ \\ $ \Rightarrow $ y=0\\
    The graph for the same is given below:

    ABCD being the feasible region.
    The Value of Z as well as the final answer is given below.
    The Value of Z at corner points A,B,C and D :
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=1000 x+500 y \\ \hline A(0,45) & Z=0+45(500)=22500 \\ \hline B(25,30) & Z=25(1000)+30(500)=4000 \rightarrow \max \\ \hline C(40,0) & Z=40(1000)+0=40000 \rightarrow \max \\ \hline D(0,0) & Z=0+0=0 \\ \hline \end{array}
    So, value of Z is maximum on-line BC, the maximum value is 40000 . So manufacturer must produce 25 number of models X and 30 number of model Y.

    Question:23

    In order to supplement daily diet, a person wishes to take some X and some wishes Y tablets. The contents of iron, calcium and vitamins in X and Y (in milligrams per tablet) are given as below:
    \begin{array}{|l|l|l|l|} \hline \text { Tablets } & \text { Iron } & \text { Calcium } & \text { Vitamin } \\ \hline \mathrm{X} & 6 & 3 & 2 \\ \hline \mathrm{Y} & 2 & 3 & 4 \\ \hline \end{array}

    The person needs at least 18 milligrams of iron, 21 milligrams of calcium and 16 milligrams of vitamins. The price of each tablet of X and Y is Rs 2 and Re 1 respectively. How many tablets of each should the person take in order to satisfy the above requirement at the minimum cost?

    Answer:

    Let us say that the number of tablet X be x and the number of tablet Y be y.
    Iron content in X and Y tablets is 6 mg and 2 mg respectively.
    Total iron content from x and y tablets = 6x + 2y
    Minimum of 18 mg of iron is required. So, we have
    \\ 6x + 2y $ \geq $ 18\\ \\ 3x + y $ \geq $ 9\\
    Similarly, calcium content in X and Y tablets is 3 mg each respectively.
    So, total calcium content from x and y tablets = 3x + 3y
    Minimum of 21 mg of calcium is required. So, we have
    \\ 6x + 2y $ \geq $ 21\\ \\ $ \Rightarrow $ x + y $ \geq $ 7\\
    Also, vitamin content in X and Y tablet is 2 mg and 4 mg respectively.
    So, total vitamin content from x and y tablets = 2x + 4y
    Minimum of 16 mg of vitamin is required. So, we have
    \\ 2x + 4y $ \geq $ 16\\ \\ $ \Rightarrow $ x + 2y $ \geq $ 8\\ \\
    Also, as number of tablets should be non-negative so, we have,
    x, y ≥ 0
    Cost of each tablet of X and Y is Rs 2 and Re 1 respectively.
    Let total cost = Z
    So, Z = 2x + y
    Finally, we have,
    Constraints,
    \\ 3x + y $ \geq $ 9\\ \\ x + y $ \geq $ 7\\ \\ x + 2y $ \geq $ 8\\ \\ x, y $ \geq $ 0\\ \\ Z = 2x + y\\
    We need to minimize Z, subject to the given constraints.
    Now let us convert the given inequalities into equation.
    We obtain the following equation
    \\ 3x + y $ \geq $ 9\\ \\ $ \Rightarrow $ 3x + y = 9\\ \\ x + y $ \geq $ 7\\ \\ $ \Rightarrow $ x + y = 7\\ \\ x + 2y $ \geq $ 8\\ \\ $ \Rightarrow $ x + 2y = 8\\ \\ x $ \geq $ 0\\ \\ $ \Rightarrow $ x=0\\ \\ y $ \geq $ 0\\ \\ $ \Rightarrow $ y=0\\ \\
    The region that is representing 3x + y ≥ 9 is the line 3x + y = 9 meets the coordinate axes (3,0) and (0,9) respectively. Once the points are joined, the lines are obtained 3x + y = 9. It does not satisfy the inequation and so the region that contains the origin then represents the solution set of the inequation 3x + y ≥ 9.
    The region that is representing x + y ≥ 7 is the line x + y = 7 meets the coordinate axes (7,0) and (0,7) respectively. Once the points are joined, the lines are obtained v It does not satisfy the inequation and so the region that contains the origin then represents the solution set of the inequation x + y ≥ 7.
    The region representing x + 2y ≥ 8 is the line x + 2y = 8 meets the coordinate axes (8,0) and (0,4) respectively. We will join these points to obtain the line x + 2y = 8. It is clear that (0,0) does not satisfy the inequation x + 2y ≥ 8. So, the region not containing the origin represents the solution set of the inequation x + 2y ≥ 8.
    The regions that represent x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.
    Given below is the graph:

    The region towards the right of ABCD is the feasible region. It is unbounded in this case.
    Value of Z at corner points A,B,C and D :
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } z=2 x+y \\ \hline A(0,9) & Z=0+9=9 \\ \hline B(1,6) & Z=1(2)+6=8 \rightarrow \text { min } \\ \hline C(6,1) & Z=6(2)+1=13 \\ \hline D(8,0) & Z=8(2)+0=16 \\ \hline \end{array}
    Now, we check if 2 x+y<8 to check if resulting open half has any point common with feasible region.
    The region represented by 2 x+y<8:
    The line 2x + y=8meets the coordinate axes (4,0) and (0,8) respectively. We will join these points to obtain the line 2x + y=8. It is clear that (0,0) satisfies the inequation 2 x+y<8. So, the region not containing the origin represents the solution set of the inequation 2 x+y<8.

    Clearly, 2x + y=8 intersects feasible region only at B.
    So, 2 x+y<8 does not have any point inside feasible region.
    So, value of Z is minimum at B(1,6), the minimum value is 8 .
    So, number of tablets that should be taken of type X and Y is 1,6 Respectively.

    Question:24

    A company makes 3 model of calculators: A, B and C at factory I and factory II. The company has orders for at least 6400 calculators of model A, 4000 calculator of model B and 4800 calculator of model C. At factory I, 50 calculators of model A, 50 of model B and 30 of model C are made every day; at factory II, 40 calculators of model A, 20 of model B and 40 of model C are made every day. It costs Rs 12000 and Rs 15000 each day to operate factory I and II, respectively. Find the number of days each factory should operate to minimise the operating costs and still meet the demand.

    Answer:

    Taking into consideration that Let number of days for which factory I operate be x and number of days for which factory II operates be y.
    Number of calculators made by factory I and II of model A are 50 and 40 respectively.
    Minimum number of calculators of model A required = 6400
    So, 50x + 40y ≥ 6400
    ⇒ 5x + 4y ≥ 640
    Number of calculators made by factory I and II of model B are 50 and 20 respectively.
    Minimum number of calculators of model B required = 4000
    So, 50x + 20y ≥ 4000
    ⇒ 5x + 2y ≥ 400
    Number of calculators made by factory I and II of model C are 30 and 40 respectively.
    Minimum number of calculators of model C requires = 4800
    So, 30x + 40y \geq 4800
    \Rightarrow 3x + 4y \geq 480
    Operating costs is Rs 12000 and Rs 15000 each day to operate factory I and II respectively.
    Let Z be total operating cost so we have Z = 12000x + 15000y
    Also, number of days are non-negative so, x, y ≥ 0
    So, we have,
    Constraints,
    \\ 5x + 4y $ \geq $ 640\\ \\ 5x + 2y $ \geq $ 400\\ \\ 3x + 4y $ \geq $ 480\\ \\ x, y $ \geq $ 0\\ \\ Z = 12000x + 15000y\\
    We need to minimize Z, subject to the given constraints.
    Now let us convert the given inequalities into equation.
    We obtain the following equation
    \\ 5x + 4y $ \geq $ 640\\ \\ $ \Rightarrow $ 5x + 4y = 640\\ \\ 5x + 2y $ \geq $ 400\\ \\ $ \Rightarrow $ 5x + 2y = 400\\ \\ 3x + 4y $ \geq $ 480\\ \\ $ \Rightarrow $ 3x + 4y = 480\\ \\ x $ \geq $ 0\\ \\ $ \Rightarrow $ x=0\\ \\ y $ \geq $ 0\\ \\ $ \Rightarrow $ y=0\\
    The region that represents the 5x + 4y ≥ 640.
    The line 5x + 4y = 640 meets the coordinate axes (128,0) and (0,160) respectively to get the final outcome. When we join these points to obtain the line 5x + 4y = 640. It is then clear that (0,0) does not satisfy the inequation 5x + 4y ≥ 640. So, the region not containing the origin represents the solution set of the inequation 5x + 4y ≥ 640.
    The region that represents the 5x + 2y ≥ 400:
    The line 5x + 2y = 400 meets the coordinate axes (80,0) and (0,200) respectively. We will join these points to obtain the line 5x + 2y = 400. It is justified that (0,0) does not satisfy the inequation 5x + 2y ≥ 400. So, the region not containing the origin represents the solution set of the inequation 5x + 2y ≥ 400.
    The region represented by 3x + 4y ≥ 480:
    The line that 3x + 4y = 480 when meets the coordinate axes (160,0) and (0,120) respectively. We will then try and join these points to obtain the line 3x + 4y = 480. It is clear that (0,0) does not justify the inequation 3x + 4y ≥ 480. So, the region not containing the origin represents the solution set of the inequation 3x + 4y ≥ 480.
    The graph is:

    The region towards the right of ABCD is the feasible region. It is unbounded in this case.
    The value of Z at the corner points, is
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=12000 \times+15000 y \\ \hline A(0,200) & Z=3000000 \\ \hline B(32,120) & Z=218400 \\ \hline C(80,60) & Z=1860000 \rightarrow \text { min } \\ \hline D(160,0) & Z=1920000 \\ \hline \end{array}
    Now, we plot 12000 x+15000 y<1860000 to check if resulting open half has any point common with feasible region.
    The region represented by 12000 \mathrm{x}+15000 \mathrm{y}<1860000
    The line 12000 \mathrm{x}+15000 \mathrm{y}=1860000 meets the coordinate axes (155,0) and (0,124) respectively. We will join these points to obtain the line $12000 \mathrm{x}+15000 \mathrm{y}=1860000. It is clear that (0,0) satisfies the inequation 12000 \mathrm{x}+15000 \mathrm{y}<1860000. So, the region containing the origin represents the solution set of the inequation 12000 \mathrm{x}+15000 \mathrm{y}<1860000.

    Clearly, 12000 \mathrm{x}+15000 \mathrm{y}<1860000 intersects feasible region only at C
    So, value of Z is minimum at C(80,60), the minimum value is 1860000 .
    So, number of days factory 1 is required to operate is 80 and number of days factory 2 should operate is 60 to minimize the cost.

    Question:25

    Maximise and Minimise Z = 3x - 4y

    subject to

    \\ x - 2y \leq 0 \\- 3x + y \leq 4 \\x - y \leq 6

    x,y \geq 0

    Answer:

    We need to maximize and minimize Z, subject to the given constraints.
    Now let us convert the given inequalities into equations.
    We obtain the following equation
    We have constraints,
    \\ x - 2y $ \leq $ 0\\ \\ - 3x + y $ \leq $ 4\\ \\ x - y $ \leq $ 6\\ \\ x, y $ \geq $ 0\\ \\ Z = 3x - 4y\\ \\
    We need to maximize and minimize Z, subject to the given constraints.
    Now let us convert the given inequalities into equations.
    We obtain the following equation
    \\x-2 y \leq 0 \\ \Rightarrow x-2 y=0 \\ -3 x+y \leq 4 \\ \Rightarrow-3 x+y=4 \\ x-y \leq 6
    \\ $ \Rightarrow $ x - y = 6\\ \\ x $ \geq $ 0\\ \\ $ \Rightarrow $ x=0\\ \\ y $ \geq $ 0\\ \\ $ \Rightarrow $ y=0\\
    The further explanation of the same is given below:
    The region represented by x – 2y ≤ 0:
    The line x - 2y = 0 meets the coordinate axes at origin and slope of the line is \frac{1}{2}. We will construct a line passing through origin and whose slope is \frac{1}{2}. As point (1,1) satisfies the inequality. So, the side of the line which contains (1,1) is feasible. Hence, the solution set of the inequation x – 2y ≤ 0 is the side which contains (1,1).
    The region represented by – 3x + y ≤ 4:
    The line – 3x + y = 4 meets the coordinate axes (-\frac{4}{3},0) and (0,4) respectively. We will join these points to obtain the line -3x + y = 4. It is clear that (0,0) satisfies the inequation – 3x + y ≤ 4. So, the region containing the origin represents the solution set of the inequation – 3x + y ≤ 4.
    The region represented by x – y ≤ 6:
    The line x – y = 6 then meets the coordinate axes (6,0) and (0,-6) respectively. We will join these points to obtain the line x – y = 6. It is clear that (0,0) satisfies the inequation x – y ≤ 6. So, the region containing the origin represents the solution set of the inequation x – y ≤ 6.
    The graph for the same is given below:


    The feasible region is region between line -3 x+y=4 \: \: and \: \: x-y=6,above BC and to the right of y axis as shown.
    Feasible region is unbounded.
    Corner points are A, B, C
    Value of Z at corner points A, B, C and D –
    \begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=3 x-4 y \\ \hline A(0,4) & Z=0-(4)(4)=-16 \rightarrow \text { min } \\ \hline B(0,0) & Z=0+0=0 \\ \hline C(12,6) & Z=3(12)-4(6)=12 \rightarrow \max \\ \hline \end{array}
    So, maximum value of Z at corner points is 12 at C and minimum is -16 at A.
    So, to check if the solution is correct, we plot 3 x-4 y>12 and 3 x-4 y<-16 for maximum and minimum respectively.
    The region represented by 3 x-4 y>12
    The line that is 3x – 4y = 12 meets the coordinate axes (4,0) and (0,-3) respectively. When we join these points, we get the line 3x – 4y > 12. It is clear that (0,0) does not satisfy the inequation 3x – 4y > 12. So, the region not containing the origin represents the solution set of the inequation 3x – 4y > 12.
    The region represented by 3x – 4y <-16:
    The line 3x – 4y = -16 meets the coordinate axes (-\frac{16}{3},0) and (0,4) respectively. We will join these points to obtain the line 3x – 4y <-16. It is clear that (0,0) does not satisfy the inequation 3x – 4y <-16. So, the region not containing the origin represents the solution set of the inequation 3x – 4y <-16.
    We get the following answer:

    Clearly, 3 x-4 y=12 has no point inside feasible region, but 3 x-4 y=-16 passes through the feasible region.
    Therefore, Z has no minimum value it has only a maximum value which is 12.

    Question:26

    The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y.

    Compare the quantity in Column A and Column B.

    \begin{array}{|l|l|} \hline \text { Column A } & \text { Column B } \\ \hline \text { Maximum of Z } & 325 \\ \hline \end{array}

    A. The quantity in column A is greater

    B. The quantity in column B is greater

    C. The two quantities are equal

    D. The relationship cannot be determined on the basis of the information supplied

    Answer:

    B)
    The quantity in column B is greater
    Z = 4x + 3y
    Corner points- (0, 0), (0, 40), (20, 40), (60, 20), (60, 0)
    A feasible region is bounded.
    Value of Z at corner points-
    At (0, 0), Z = 0
    At (0, 40), Z = 120
    At (20, 40), Z = 200
    At (60, 20), Z = 300
    At (60, 0), Z = 240
    Clearly, the maximum value of Z is 300, which is less than 325.

    Question:27

    The feasible solution for a LPP is shown in Fig. 12.12. Let Z = 3x – 4y be the

    objective function. Minimum of Z occurs at

    A. (0, 0)

    B. (0, 8)

    C. (5, 0)

    D. (4, 10)

    Answer:

    B)
    Answer
    Value of Z = 3x – 4y, at corner points are –
    At (0, 0) = 0
    At (0, 8) = -32
    At (4, 10) = -28
    At (6, 8) = -14
    At (6, 5) = -2
    At (5, 0) = 15
    So, clearly minimum value is at (0, 8)

    Question:28

    Refer to Exercise 27. Maximum of Z occurs at

    A. (5, 0)

    B. (6, 5)

    C. (6, 8)

    D. (4, 10)

    Answer:

    Correct Answer A
    Answer
    Value of Z = 3x – 4y, at corner points are –
    At (0, 0) = 0
    At (0, 8) = -32
    At (4, 10) = -28
    At (6, 8) = -14
    At (6, 5) = -2
    At (5, 0) = 15
    So, clearly maximum value is at (5, 0)

    Question:29

    Refer to Exercise 27. (Maximum value of Z + Minimum value of Z) is equal to

    A. 13

    B. 1

    C. – 13

    D. – 17

    Answer:

    Correct Answer D)
    Answer:
    Value of Z = 3x – 4y, at corner points are –
    At (0, 0) = 0
    At (0, 8) = -32
    At (4, 10) = -28
    At (6, 8) = -14
    At (6, 5) = -2
    At (5, 0) = 15
    So, clearly maximum value is at (5, 0)
    Maximum value = 15,
    Minimum value = -32
    So, maximum + minimum = 15 -32
    = -17

    Question:30

    The feasible region for an LPP is shown in the Fig. 12.13. Let F = 3x – 4y be the objective function. Maximum value of F is.

    A. 0
    B. 8
    C. 12
    D. – 18

    Answer:

    Correct Answer C)
    Answer:
    F = 3x – 4y
    Corner points are (0, 0), (0, 4), (12, 6)
    Value of F at corner points –
    At (0, 0), F = 0
    At (0, 4), F = -16
    At (12, 6), F = 12
    Clearly, the maximum value of F = 12.

    Question:31

    Refer to Exercise 30. Minimum value of F is
    A. 0
    B. – 16
    C. 12
    D. does not exist

    Answer:

    Correct Answer B)
    Answer:
    F = 3x – 4y
    Corner points are (0, 0), (0, 4), (12, 6)
    Value of F at corner points –
    At (0, 0), F = 0
    At (0, 4), F = -16
    At (12, 6), F = 12
    Clearly, the minimum value of F = – 16.

    Question:32

    Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y be the objective function.
    The Minimum value of F occurs at

    A. (0, 2) only
    B. (3, 0) only
    C. the midpoint of the line segment joining the points (0, 2) and (3, 0) only
    D. any point on the line segment joining the points (0, 2) and (3, 0).

    Answer:

    Correct Answer D)
    Answer:
    F = 4x + 6y

    Corner points - (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).
    Value of F at corner points –
    At (0, 2), F = 12
    At (3, 0), F = 12
    At (6, 0), F = 24
    At (6, 8), F = 72
    At (0, 5), F = 30
    Feasible region –
    As, feasible region to be bounded so it is a closed polygon.
    So, minimum values of F = 12 are at (3, 0) and (0, 2).
    Therefore, the minimum value of F occurs at any point on the line segment joining the points (0, 2) and (3, 0).

    Question:33

    Refer to Exercise 32, Maximum of F – Minimum of F =
    A. 60
    B. 48
    C. 42
    D. 18

    Answer:

    Correct Answer A)
    Answer:
    F = 4x + 6y
    Corner points - (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).
    Value of F at corner points –
    At (0, 2), F = 12
    At (3, 0), F = 12
    At (6, 0), F = 24
    At (6, 8), F = 72
    At (0, 5), F = 30
    Feasible region –

    Considering, feasible region to be bounded so it is a closed polygon.
    Minimum value of F = 12
    Maximum value of F = 72
    So, Maximum of F – Minimum of F = 60.

    Question:34

    Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px+qy , where p, q > 0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is
    A. p = 2q
    B. p = q/2
    C. p = 3q
    D. p = q

    Answer:

    Given, Z = px+qy
    Given, minimum occurs at (3, 0) and (1, 1).
    For a minimum to occur at two points the value of Z at both points should be the same.
    So, value of Z at (3, 0) = value of Z at (1, 1)
    ⇒ 3p = p + q
    ⇒ 2p = q
    \Rightarrow p=\frac{q}{2}

    So, option B is correct.

    Question:35

    Fill in the blanks in each of the.
    In a LPP, the linear inequalities or restrictions on the variables are called _________.

    Answer:

    In a LPP, the linear inequalities or restrictions on the variables are called linear constraints.

    Question:36

    Fill in the blanks in each of the Exercise.
    In a LPP, the objective function is always _________

    Answer:

    In a LPP, the objective function is always linear.

    Question:37

    Fill in the blanks in each of the.
    If the feasible region for a LPP is _________, then the optimal value of the objective function Z = ax + by may or may not exist.

    Answer:

    If the feasible region for a LPP is unbounded, then the optimal value of the objective function Z = ax + by may or may not exist.

    Question:38

    Fill in the blanks in each of the.
    In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same _________ value.

    Answer:

    In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same maximum value

    Question:39

    Fill in the blanks in each of the.
    A feasible region of a system of linear inequalities is said to be _________ if it can be enclosed within a circle.

    Answer:

    A feasible region of a system of linear inequalities is said to be bounded if it can be enclosed within a circle.

    Question:40

    Fill in the blanks in each of the Exercise.
    A corner point of a feasible region is a point in the region which is the _________ of two boundary lines.

    Answer:

    A corner point of a feasible region is a point in the region which is the intersection of two boundary lines.

    Question:41

    Fill in the blanks in each of the Exercise.
    The feasible region for an LPP is always a _________ polygon.

    Answer:

    The feasible region for an LPP is always a convex polygon.

    Question:42

    State whether the statements in Exercise are True or False.
    If the feasible region for a LPP is unbounded, the maximum or minimum of the objective function Z = ax + by may or may not exist.

    Answer:

    True
    If the feasible region is unbounded then we may or may not have a maximum or minimum of objective function, but if we have a maximum or a minimum value then it must be at one of the corner points only.

    Question:43

    State whether the statements in Exercise are True or False.
    Maximum value of the objective function Z = ax + by in a LPP always occurs at only one corner point of the feasible region.

    Answer:

    False.
    Maximum value or minimum value can occur at more than one point. In such all the points lie on a line segment and are part of the boundary of the feasible region.

    Question:44

    State whether the statements in Exercise are True or False.
    In a LPP, the minimum value of the objective function Z = ax + by is always 0 if origin is one of the corner points of the feasible region.

    Answer:

    False.
    Minimum value of objective function can also be negative if the coefficient of x or y is negative. So, it is not necessary that the minimum value of objective function will be zero.

    Question:45

    State whether the statements in Exercise are True or False.
    In a LPP, the maximum value of the objective function Z = ax + by is always finite.

    Answer:

    False.
    In a LPP, the maximum value of the objective function Z = ax + by may or may not be finite. It depends on the feasible region. If a feasible region is unbounded then we can also have infinite maximum value of objective function.

    More About NCERT Exemplar Solutions for Class 12 Maths Chapter 12

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    Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 12

    • Some of the broad applications of Linear Programming include areas of food and agriculture, engineering, transportation, manufacturing, and energy. By determining the number of crops and their efficient use, farmers can make the best use of their resources and gather revenue. Engineers often use these problems to help with aerodynamic shape optimisation. The scheduling, travel time, and passengers in the transportation industry are done with Linear Programming. In times of war, linear programming concepts were used to deal with the efficient distributing of resources, keeping in mind the various restrictions such as prices and availability.
    • Linear Programming can also be applied to various diet problems, from food aid, national food programs, and dietary guidelines to individual issues after applying certain nutritional constraints, costs constraints, acceptability constraints, and ecological constraints. A student familiar with these applications would have sound concepts and excel in his/her exams.

    NCERT Exemplar Class 12 Solutions

    Also, check NCERT Solutions for questions given in the book:

    Must Read NCERT Solution subject wise

    Read more NCERT Notes subject wise

    Also Check NCERT Books and NCERT Syllabus here

    Frequently Asked Question (FAQs)

    1. How is linear programming an important chapter in maths of Class 12?

    These days linear programming is helpful in almost every field and industry from finance to engineering to medicine. Learning this chapter will help in getting a better understanding of mathematical modeling.

    2. Are these solutions helpful in entrance exam preparation?

    Yes, these NCERT exemplar Class 12 Maths solutions chapter 12 are helpful in the preparation of all types of engineering entrance exams like JEE Main and others.

    3. Who prepares these NCERT Class 12 chapter 12 solutions of maths?

    These solutions are prepared by our highly experienced experts and teachers, who have years of experience and knowledge regarding NCERT and CBSE pattern and requirement.

    4. Are these NCERT solutions exhaustive?

    Yes, the NCERT exemplar solutions for Class 12 Maths chapter 12 given here are exhaustive with each step mentioned in complete detail. The solutions are also as per the CBSE guidelines and pattern.

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    Questions related to CBSE Class 12th

    Have a question related to CBSE Class 12th ?

    hello,

    Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

    I hope this was helpful!

    Good Luck

    Hello dear,

    If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


    As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


    Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


    Believe in Yourself! You can make anything happen


    All the very best.

    Hello Student,

    I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

    You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

    All the best.

    If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

    Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

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    A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

    Option 1)

    0.34\; J

    Option 2)

    0.16\; J

    Option 3)

    1.00\; J

    Option 4)

    0.67\; J

    A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

    Option 1)

    2.45×10−3 kg

    Option 2)

     6.45×10−3 kg

    Option 3)

     9.89×10−3 kg

    Option 4)

    12.89×10−3 kg

     

    An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

    Option 1)

    2,000 \; J - 5,000\; J

    Option 2)

    200 \, \, J - 500 \, \, J

    Option 3)

    2\times 10^{5}J-3\times 10^{5}J

    Option 4)

    20,000 \, \, J - 50,000 \, \, J

    A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

    Option 1)

    K/2\,

    Option 2)

    \; K\;

    Option 3)

    zero\;

    Option 4)

    K/4

    In the reaction,

    2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

    Option 1)

    11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

    Option 2)

    6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

    Option 3)

    33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

    Option 4)

    67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

    How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

    Option 1)

    0.02

    Option 2)

    3.125 × 10-2

    Option 3)

    1.25 × 10-2

    Option 4)

    2.5 × 10-2

    If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

    Option 1)

    decrease twice

    Option 2)

    increase two fold

    Option 3)

    remain unchanged

    Option 4)

    be a function of the molecular mass of the substance.

    With increase of temperature, which of these changes?

    Option 1)

    Molality

    Option 2)

    Weight fraction of solute

    Option 3)

    Fraction of solute present in water

    Option 4)

    Mole fraction.

    Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

    Option 1)

    twice that in 60 g carbon

    Option 2)

    6.023 × 1022

    Option 3)

    half that in 8 g He

    Option 4)

    558.5 × 6.023 × 1023

    A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

    Option 1)

    less than 3

    Option 2)

    more than 3 but less than 6

    Option 3)

    more than 6 but less than 9

    Option 4)

    more than 9

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    Also Read: Career as Nurse

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    Resource Links for Online MBA 

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    Procurement Manager

    The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness. 

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    Production Planner

    Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner. 

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    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    3 Jobs Available
    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    IT Consultant

    An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

    2 Jobs Available
    Data Architect

    A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements. 

    2 Jobs Available
    Security Engineer

    The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

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    UX Architect

    A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy. 

     

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