RD Sharma Class 12 Exercise CSBQ Maxima And Minima Solutions Maths

RD Sharma Class 12 Exercise CSBQ Maxima And Minima Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 21, 2022 02:48 PM IST

The Class 12 RD Sharma chapter 17 exercise CSBQ solution plays a major role in making the students understand the mathematical concepts to a great extent. The class 12 students suffer due to lack of proper guidance to learn mathematics. The presence of the RD Sharma class 12th exercise CSBQ eradicates that emptiness and gives a strong support system to the students.

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RD Sharma Class 12 Solutions Chapter 17 CSBQ Maxima and Minima - Other Exercise
Maxima and Minima Excercise: CSBQ
RD Sharma Chapter-wise Solutions

Also read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 17 CSBQ Maxima and Minima - Other Exercise

Maxima and Minima Excercise: CSBQ

Maxima and Minima Exercise Case Study Based Questions Question 1(i)

Answer:

2 x + π y = 200.

Hint: perimeter = sum of all sides.
Solution:
Give the perimeter of the region=200m

x + \frac{1}{2} \times 2 π r + x + \frac{1}{2} \times 2 π r = 200

2 x + 2 π r = 200

Putting

2 x + 2 π (\frac{y}{2}) = 200

2 x + π y = 200.

Maxima and Minima Exercise Case Study Based Questions Question 1(ii)

Answer:

\frac{2}{π} (100 x - x^{2})

Hint: Area of rectangle = length × breadth
Solution:
Area =Area of rectangle

= x \times y

$\ since, 2x+ \pi y =200$
$\ \pi y =200-2x$

y = \frac{1}{π} (200 - 2 x)

Area =

x \times \frac{1}{π} (200 - 2 x)

Area =

\frac{1}{π} (200 x - 2 x^{2})

= \frac{2}{π} (100 x - x^{2})

Maxima and Minima Exercise Case Study Based Questions question 1(iii)

Answer:

\frac{5000}{π} m^{2}

Hint: use maxima minima concept.
Solution:
Now,

A = \frac{2}{π} (100 x - x^{2})

To find maximum value of A we find

\frac{dA}{dx}

We must show f’(x) = 0 for finding maxima and minima

\begin{aligned} \frac{d A}{d x} = \frac{2}{π} (100 - 2 x) \end{aligned}

\frac{d A}{d x} = 0

\frac{2}{π} (100 - 2 x) = 0

100 = 2 x

x = 50

Putting x value in A

\begin{aligned} A & = \frac{2}{π} (100 x - x^{2}) \end{aligned}

= \frac{2}{π} (100 \times 50 - (50)^{2})

= \frac{2}{T} (5000 - 2500)

A . = \frac{2}{π} (2500) = \frac{5000 m^{2}}{π}

Maxima and Minima Exercise Case Study Based Questions Question 1(iv)

Answer:

x = 0

Hint: use concept of maxima and minima.
Solution:
Total Area = =Area of rectangle+2× Area of semicircle.

= A + 2 \times \frac{1}{2} π r^{2}

\begin{aligned} = A + π r^{2} \end{aligned}

Putting

r = \frac{y}{2}

= A + π (y / 2)^{2}

\begin{aligned} = A + \frac{π y^{2}}{4} \end{aligned}

Putting A value,

= \frac{2}{π} (100 x - x^{2}) + \frac{π y^{2}}{4}

Putting

y = \frac{1}{π} (200 - 2 x)

\begin{aligned} = \frac{2}{π} (100 x - x^{2}) + \frac{π}{4} {(\frac{1}{π} (200 - 2 x))}^{2} \end{aligned}

= \frac{2}{π} (100 x - x^{2}) + \frac{π}{4} \times \frac{1}{π^{2}} (200 - 2 x)^{2}

\begin{aligned} = \frac{2}{π} (100 x - x^{2}) + \frac{π}{4} \times \frac{1}{π^{2}} 2^{2} \times (100 - x)^{2} \end{aligned}

= \frac{200 x}{π} - \frac{2 x^{2}}{π} + \frac{100^{2}}{π} + \frac{x^{2}}{π} - \frac{200 x}{π}

\begin{aligned} = \frac{- x^{2}}{π} + \frac{100^{2}}{π} \end{aligned}

Total Area

= \frac{- x^{2}}{π} + \frac{100^{2}}{π}

Let Total Area =Z
$z=\frac{-x^{2}}{\pi}+\frac{10000}{\pi} $

\begin{aligned} \frac{d z}{d x} = \frac{- 2 x}{π} \end{aligned}

\frac{d z}{d x} = 0

\frac{- 2 x}{π} = 0

\begin{aligned} x = 0 \end{aligned}

Maxima and Minima Excercise Case Study Based Questions Question 1(v)

Answer:

\frac{5000 m^{2}}{π}

Hint: use concept of maxima and minima.
Solution:
Total area

Z = \frac{- x^{2}}{π} + \frac{10000}{π}

Putting,

x = 0

\begin{aligned} Z = \frac{100^{2}}{π} \end{aligned}

Z = \frac{10000}{π} m^{2}

Also Maximum area,

A = \frac{5000}{π} m^{2}

Thus,
Difference between areas

\begin{aligned} = \frac{10000}{π} - \frac{5000}{π} \end{aligned}

= \frac{5000 m^{2}}{π}

Maxima and Minima Excercise Case Study Based Questions Question 2(i)

Answer:

0.04

Solution:
Let A: Vinay process the form
Let B: Sonia process the form
Let A: Iqbal process the form
We need to find
The conditional probability that an error is committed in processing given that Sonia processed the form is

P (\frac{E}{B})

Since, Sonia’s error rate

0.04

P (\frac{E}{B}) = 0.04

P(E/B)=0.04

Maxima and Minima Excercise Case Study Based Questions Question 2(iii)

Answer: $0.04 F$
Solution:
The total probability of committing an error = Probability Vinay processes the form × Vinay’s error rate + Probability Sonia processes the form × Sonia’s error rate + Probability Iqbal processes the form × Iqbal’s error rate

\begin{aligned} = 50 % \times 0.06 + 20 % \times 0.04 + 30 % \times 0.03 \end{aligned}

= \frac{5}{10} \times 0.06 + \frac{2}{10} \times 0.04 + \frac{3}{10} \times 0.03

\begin{aligned} = 0.030 + 0.008 + 0.009 \end{aligned}

= 0.047

Maxima and Minima Exercise Case Study Based Questions Question 2(iv)

Answer: $3047$
Solution:
We first find the probability that an error is committed in processing given that Vinay processed the form

P (\frac{A}{E})

Now,

\begin{aligned} P (\frac{A}{E}) & = \frac{P (A) \cdot P (\frac{E}{A})}{p (A) \cdot p (\frac{E}{A}) + P (B) \cdot P (\frac{E}{B}) + P (C) - P (\frac{E}{C})} \end{aligned}

= \frac{5}{\frac{5}{10} \times 0.06 + \frac{2}{10} \times 0.44 + \frac{3}{10} \times 0.03}

\begin{aligned} = \frac{5 \times 0.06}{5 \times 0.06 + 2 \times 0.04 + 3 \times 0.03} \end{aligned}

= \frac{0.30}{0.30 + 0.08 + 0.19}

= \frac{0.30}{0.47}

= \frac{30}{47}

Probability form has an error but was not processed by Vinay,

\begin{aligned} = 1 - P (\frac{A}{E}) \end{aligned}

= 1 - \frac{30}{47}

= \frac{17}{47}

Maxima and Minima Exercise Case Study Based Questions Question 2(v)

Answer: $1$
Solution:

\sum_{i = 1}^{3} P (\frac{E_{i}}{A}) = p (\frac{E_{1}}{A}) + P_{2} (\frac{E_{2}}{A}) + p (\frac{E_{3}}{A})

= \frac{P (E_{1}) \cdot P (\frac{A}{E_{1}})}{P (E_{1}) \cdot P (\frac{A}{E_{1}}) + P (E_{2}) \cdot P (\frac{A}{E_{2}}) + P (E_{3}) \cdot P (\frac{A}{E_{3}})} + \frac{P (E_{2}) \cdot P (\frac{A}{E_{2}})}{P (E_{1}) \cdot P (\frac{A}{E_{1}}) + P (E_{2}) \cdot P (\frac{A}{E_{2}}) + P (E_{3}) \cdot P (\frac{A}{E_{3}})} +

\begin{aligned} \frac{P (E_{3}) \cdot P (\frac{A}{E_{3}})}{P (E_{1}) \cdot P (\frac{A}{E_{1}}) + P (E_{2}) \cdot P (\frac{A}{E_{2}}) + P (E_{3}) \cdot P (\frac{A}{E_{3}})} \end{aligned}

= 1

Maxima and Minima Exercise Case Study Based Questions Question 3(i)

Answer: $R = Rs (500 - x) (300 + x)$
Hint: use the concept of revenue.
Solution:
Total Revenue R = No. of Subscribers × Rate

= (500 - x) \times (300 + x)

Where x is the Rs for increasing one subscriber.

Maxima and Minima Excercise Case Study Based Questions Question 3(ii)

Answer:

AR = R s (\frac{500}{x} - 1) (300 + x)

Hint: use concept of revenue.
Solution:
Normal Revenue is

R = (500 - x) (300 + x)

The average revenue of the company when annual subscription is increased by Rs x per subscibers
Average revenue

\begin{aligned} = (\frac{500 - x}{x}) (300 + x) \\ = Rs (\frac{500}{x} - 1) (300 + x) \end{aligned}

Maxima and Minima Excercise Case Study Based Questions Question 3(iii)

Answer:

200 - 2 x

Hint: use concept of maxima and minima.
Solution:

\begin{aligned} R (x) = (500 - x) (300 + x) \end{aligned}

R (x) = - x^{2} + 200 x + 150000

Differentiating with respect to x,

\begin{aligned} R^{'} (x) = - 2 x + 200 \end{aligned}

Hence = 200 - 2 x

Maxima and Minima Excercise Case Study Based Questions Question 3(iv)

Answer:

Rs 400

Hint: use concept of maxima and minima.
Solution:

\begin{aligned} R & = (500 - x) (300 + x) \end{aligned}

R^{'} = - (300 + x) + 500 - x

= 200 - 2 x

R^{'} = 0

200 - 2 x = 0

x = 100 .

R^{''} = - 2 < 0 Maximum.

Rate = 300 + 100 = 400 R s

Maxima and Minima Excercise Case Study Based Questions Question 3(v)

Answer:

160, 000

Hint: use concept of maxima and minima.
Solution:
Given that there are 5000 subscribers and each subscriber pays 3000 Rs as rent per year
Let us assume that x rupees are increased in the rent.
So the total cost received by the company C is

\begin{aligned} C = (3000 + x) (5000 - x) \\ C = 15000000 + 2000 x - x^{2} \end{aligned}

Differentiate it with respect to x, We get

\begin{aligned} \frac{dc}{dx} = 2000 - 2 x \end{aligned}

2000 - 2 x = 0

x = 1000

So the maximum cost required by the company is

4000 \times 4000 = 160, 000

Maxmima and Minima Exercise Case Study Questions Question 4(i)

Answer:

2 x + 2 y + π x = 10

Hint: use the formula of perimeter.
Solution:
Perimeter of the window = 10 m
Let 2x, y represents width and height
Length +2× breadth + circumference of semicircle = 10

\begin{aligned} 2 x + 2 y + π (\frac{2 x}{2}) = 10 \end{aligned}

2 x + 2 y + π x = 10

Maxmima and Minima Exercise Case Study Questions Question 4(ii)

Answer: $10 x - (π + 2) x^{2}$
Hint: use the formulas of area and perimeter.
Solution:
Perimeter of the window is given, by

P = 2 x + 2 y + π x

\begin{aligned} p = 2 x + 4 y + π x \end{aligned}

(∴ p = 10 m)

\begin{aligned} 2 x + 4 y + π x & = 10 \end{aligned}

y = \frac{10 - π x - 2 x}{4} - (1)

Area = (2 x) (2 y) + \frac{π x^{2}}{2}

[∴Area of the window = Area of the Rectangle + Area of the Semicircle]

\begin{aligned} A = 4 x y + \frac{π x^{2}}{2} - - - (2) \end{aligned}

y value in (2)

\begin{aligned} A = 4 x (\frac{10 - π x - 2 x}{4}) + \frac{π x^{2}}{2} \end{aligned}

= 10 x - π x^{2} - 2 x^{2} + \frac{π x^{2}}{2}

\begin{aligned} = 10 x - 2 x^{2} - \frac{π x^{2}}{2} \end{aligned}

= 10 x - \frac{4 x^{2} + π x^{2}}{2}

= 10 x - (2 x^{2} + π x^{2})

A = 10 x - x^{2} (π + 2)

Maxmima and Minima Exercise Case Study Questions Question 4(iii)

Answer:

10 x - (\frac{π + 4}{2}) x^{2}

Hint: use the basic concept.
Solution:

\begin{aligned} Area & = (2 x) (2 y) + \frac{π x^{2}}{2} \end{aligned}

= 4 x (\frac{10 - π x - 2 x}{4}) + \frac{π x^{2}}{2}

\begin{aligned} A & = 10 x - 2 x^{2} - \frac{π x^{\land} 2}{2} \end{aligned}

= 10 x - \frac{4 x^{2} + π x^{2}}{2}

A = 10 x - x^{2} (\frac{π + 4}{2})

Maxmima and Minima Exercise Case Study Questions Question 4(iv)

Answer: $\frac{20}{π + 4}, \frac{10}{π + 4}$
Hint: use concept of maxima and minima.
Solution:
Let x, y be the length and breadth of the rectangle ABCD, Let

\frac{x}{2}

be the radius of semicircle with center at O
Perimeter = 10 m

x + 2 y + \frac{π x}{2} = 10

2 x + 4 y + π x = 20

$4 y=20-(\pi+2) x $

y = \frac{20 - (π + 2) x}{4}

\begin{aligned} A = x y + \frac{1}{2} π {(\frac{x}{2})}^{2} \end{aligned}

= x [\frac{20 - (π + 2) x}{4}] + [\frac{x^{2} π}{8}]

A = 1 / 4 [20 x - (π + 2) x^{2}] + [\frac{π x^{2}}{8}]

\begin{aligned} \frac{d A}{d x} = \frac{1}{4} [20 - 2 (π + 2) x] + \frac{π x}{4} \end{aligned}

\begin{aligned} \frac{d A}{dx} & = 0 \end{aligned}

\frac{1}{4} [20 - 2 (π + 2) x] + \frac{π x}{4} = 0 .

20 - 2 π x - 4 x + π x = 0

20 - (π + 4) x = 0

\begin{aligned} x = \frac{20}{π + 4} \end{aligned}

And,
$y=\frac{20-(\pi+2)\left(\frac{20}{\pi+4}\right)}{4} $

\begin{aligned} = \frac{- 20 π + 80 - 20 π - 40}{4 (π + 4)} \end{aligned}

y = \frac{10}{π + 4}

Maxmima and Minima Exercise Case Study Questions Question 4(v)

Answer:

\frac{50}{π + 4}

Hint: use formula of area of semicircle.
Solution:
Area of maximum length throughout the window

= \frac{radius of Semicircle (x)}{2} + breadth of rectangle

\begin{aligned} = (\frac{20}{\frac{π + 4}{2}}) + \frac{10}{π + 4} \cdot \end{aligned}

= \frac{40}{π + 4} + \frac{10}{π + 4} = \frac{50}{π + 4}

Maxima and Minima Exercise Case Study Based Questions Question 5(i)

Answer: $1800xy$
Hint: use concept.
Solution:
X is length, y is breadth
Area = l × b

A = xy

, For one square meter the cost is 1800
So,

A = 1800 x y

Maxima and Minima Exercise Case Study Based Questions Question 5(ii)

Answer:
Hint: perimeter of rectangle

\begin{aligned} = 2 (l \times b) \end{aligned}

Solution:
The cost of construction of the vertical wall is like Perimeter
Perimeter

\begin{aligned} = 2 (l \times b) \end{aligned}

= 2 (x + y)

For one square meter the cost is 1800
Then the cost of a vertical wall is

\begin{aligned} = 2 \times 1800 (x + y) \end{aligned}

= 3600 (x + y)

Maxima and Minima Exercise Case Study Based Questions Question 5(iii)

Answer:

3600 (1 - \frac{4}{x^{2}})

Hint: volume

= l \times b \times h

Solution:
Let c be the total cost of the tank
C(x) = Cost of base + Cost of sides

= 1800 xy + 3600 (x + y) \to (1)

As w know,
Length if the tank

= x

Breadth of the tank

= y

Depth of the tank

= h

= 2 m

Volume of the tank

\begin{aligned} = 8 m^{3} \end{aligned}

\begin{aligned} 1 = x \\ b = y \\ h = 2 \\ v = 8 m^{2} \end{aligned}

Volume

= l \times b \times h

\begin{aligned} 8 = 2 \times x \times y \\ y = \frac{4}{x} \Rightarrow (2) \end{aligned}

Sub y value in (1)

\begin{aligned} C (x) & = 1800 \times (\frac{4}{x}) + 3600 (x + 4) \end{aligned}

\Rightarrow 7200 + 3600 (x + 4 x^{- 1})

c^{'} (x) = 0 + 3600 (1 + (- 1) 4 x^{- 1 - 1})

= 3600 (1 - 4 x^{- 2})

= 3600 (1 - \frac{4}{x^{2}})

Maxima and Minima Exercise Case Study Based Questions Question 5(iv)

Answer:

x = 2

Hint: use the concept of maxima and minima.
Solution:
As we know,

C^{'} (x) = 3600 (1 - \frac{4}{x^{2}})

Putting

c^{'} (x) = 0

\begin{aligned} 3600 (1 - \frac{4}{x^{2}}) & = 0 . \end{aligned}

1 - \frac{4}{x^{2}} = 0

\frac{x^{2} - 4}{x^{2}} = 0

x^{2} - 4 = 0

(x - 2) (x + 2) = 0

\begin{aligned} x = 2 \end{aligned}

c^{''} (x) = 3600 (0 - 4 (- 2) x^{- 2 - 1})

= 3600 (0 - 4 (- 2) x^{- 3}

= 3600 (8 x^{- 3})

= 28800 x^{- 3}

= \frac{28800}{x^{3}}

Putting

\begin{aligned} x = 2 \end{aligned}

\begin{aligned} c^{''} (2) = \frac{28800 > 0}{(2)^{3}} \\ c^{''} (2) > 0 \end{aligned}

So minimum is

\begin{aligned} x = 2 \end{aligned}

Maxima and Minima Exercise Case Study Based Questions Question 5(v)

Answer: 21600
Hint: use basic concepts.
Solution:
Least cost of construction is = C (2)

\begin{aligned} C (2) & = 7200 + 3600 (x + 4 x^{- 1}) \end{aligned}

= 7200 + 3600 (2 + \frac{4}{2})

\begin{aligned} = 7200 + 3600 (\frac{4 + 4}{2}) \end{aligned}

= 7200 + 3600 (4)

= 7200 + 14400

= 21, 600

Maxima and Minima Excercise Case Study Based Questions question 6(i)

Answer: $2 x + π y = 14$
Hint: use formula of perimeter.
Solution:
Wire of Length is

= 28 m

Side of the Square

= x

Circle of Radius

= y

Sum of perimeter of a circle is

= 28 m

\begin{aligned} 2 π (Radius) + 4 (side) = 28 \end{aligned}

2 π y + 4 x = 28

2 (π y + 2 x) = 28

2 x + π y = 14

Maxima and Minima Excercise Case Study Based Questions Question 6(ii)

Answer:

\frac{1}{π} [(π + 4) x^{2} - 56 x + 196]

Hint: use formulas.
Solution:

\begin{aligned} Area = π (raduis)^{2} + (side)^{2} \end{aligned}

A = π y^{2} + x^{2} \to (1)

We know that

\begin{aligned} 2 x + π y & = 14 \end{aligned}

π y = 14 - 2 x

y = \frac{14 - 2 x}{π}

Substituting value of

y in (1) π {(\frac{14 - 2 x}{π})}^{2} + \frac{x^{2}}{4^{2}}

\begin{aligned} A & = π {(\frac{14 - 2 x}{π})}^{2} + \frac{x^{2}}{16} \end{aligned}

= \frac{1}{π} (14 - 2 x)^{2} + \frac{x^{2}}{16}

= \frac{1}{π} (14^{2} - 2 (14) (2 x) + (2 x)^{2}) + \frac{x^{2}}{16} .

\begin{aligned} = \frac{1}{π} [196 - 56 x + 4 x^{2}] + \frac{x^{2}}{16} \end{aligned}

= \frac{1}{π} [(π + 4) x^{2} - 56 x + 196]

Maxima and Minima Excercise Case Study Based Questions Question 6(iii)

Answer:

\frac{28 π}{4 + π}

Hint: use the concept of maxima and minima.
Solution:
Side of the Square

= \frac{28 - x}{4}

Area of the square

= {[\frac{28 - x}{4}]}^{2}

Thus Total Area

= \frac{x^{2}}{4 π} {[\frac{28 - x}{4}]}^{2}

\begin{aligned} \frac{d d}{d x} = \frac{2 x}{4 π} + \frac{2}{16} (28 - x) (- 1) \end{aligned}

\frac{d d}{d x} = \frac{x}{2 π} - \frac{28 - x}{8}

\begin{aligned} \frac{dA}{dx} = 0 \end{aligned}

\frac{x}{2 π} - \frac{28 - x}{8} = 0

4 x = 28 π - π x

4 x + π x = 28 π

\begin{aligned} x [4 + π] = 28 π \end{aligned}

x = \frac{28 π}{4 + π}

Maxima and Minima Excercise Case Study Based Questions Question 6(iv)

Answer:

\frac{14 π}{π + 4}

Hint: use basics.
Solution:

2 x + π y = 14

As we know,

\begin{aligned} x = \frac{28 π}{4 + π} \end{aligned}

= 2 (\frac{28 π}{4 + π}) + π y = 14

\begin{aligned} (\frac{56 π}{4 + π}) + π y = 14 \end{aligned}

\frac{56 π}{4 + π} = 14 - π y

\begin{aligned} - 14 + \frac{56 π}{4 + π} & = - π y \end{aligned}

- 56 π - 14 π + 56 π = - π y (4 + π) .

- \frac{14 π}{π + 4} = - y

\frac{14 π}{π + 4} = y

Maxima and Minima Excercise Case Study Based Questions Question 6(v)

Answer:

\frac{196 π - 2}{π (π + 4)}

Hint: use basic concepts and concepts of maxima and minima.
Solution:

\begin{aligned} A = π {(\frac{14 - 2 x}{4})}^{2} + \frac{x^{2}}{4^{2}} \end{aligned}

= π {(\frac{14 - 2 x}{4})}^{2} + \frac{x^{2}}{16}

= \frac{π (196 + 4 x^{2} - 56 x)}{16} + \frac{x^{2}}{16}

Put

x = - 1

\begin{aligned} A & = \frac{π (196) - 2}{π (π + 4)} \end{aligned}

= \frac{196 π - 2}{π (π + 4)}

Maxmima and Minima Exercise Case Study Questions Question 7(i)

Answer:

120 - \sqrt{x^{2} - 25600}

Hint: use distance formula.
Solution:

GA = GB = x

Let G be the position of the godown at a distance x each from A and B

\begin{aligned} C D = \sqrt{200^{2} - 160^{2}} \end{aligned}

= 120

G D = \sqrt{x^{2} - 160^{2}}

∴ G C = D C - D G

\begin{aligned} = 120 - \sqrt{x^{2} - 160^{2}} \end{aligned}

= 120 - \sqrt{x^{2} - 25600}

Maxmima and Minima Exercise Case Study Questions Question 7(ii)

Answer:

2 - \frac{x}{\sqrt{x^{2} - 25600}}

Hint: use basic concepts.
Solution:
We know,

y = GA + GB + GC

Then,

\begin{aligned} y = 2 x + 120 - \sqrt{x^{2} - 160^{2}} \end{aligned}

\frac{d y}{d x} = 2 - \frac{1}{2 \sqrt{x^{2} - 160^{2}}}

\frac{d y}{d x} = 2 - \frac{x}{\sqrt{x^{2} - 25600}}

Maxmima and Minima Exercise Case Study Questions Question 7(iii)

Answer:

\frac{25600}{{(x^{2} - 25600)}^{3 / 2}}

Hint: use the concept of maxima and minima.
Solution:

\begin{aligned} \frac{d y}{d x} & = 2 - \frac{x}{\sqrt{x^{2} - 25600}} \end{aligned}

\begin{aligned} \frac{d^{2} y}{d x^{2}} & = - [\frac{1}{\sqrt{(x^{2} - 25600)}} + x \cdot \frac{- 1}{2 {(x^{2} - 25600)}^{\frac{3}{2}}} \cdot 2 x] \end{aligned}

= [\frac{25600}{{(x^{2} - 25600)}^{3 / 2}}]

Maxmima and Minima Exercise Case Study Questions Question 7(iv)

Answer:

x = \frac{320}{\sqrt{3}}

Hint: use the concept of maxima and minima.
Solution:

\begin{aligned} \frac{d^{2} y}{d x^{2}} = [\frac{25600}{{(x^{2} - 25600)}^{5 / 2}}] \end{aligned}

\frac{d y}{d x} = 2 - \frac{1}{2 \sqrt{{(x^{2} - 100)}^{2}}} \cdot 2 x = 0

For max or min

\begin{aligned} \frac{2}{x} = \frac{1}{\sqrt{x^{2} - (160)^{2}}} \to (1) \end{aligned}

\frac{d^{2} y}{d x^{2}} = - [\frac{2}{x} - x^{2} \cdot \frac{8}{x^{3}}] by (1) = \frac{6}{x}

y is minimum when

x = \frac{320}{\sqrt{3}}

Maxmima and Minima Exercise Case Study Questions question 7(v)

Answer: $40 (4 \sqrt{3} + 3)$
Hint: use basic concept.
Solution:

\begin{aligned} GA + GB + GC = ? \end{aligned}

GB = GA = x

\Rightarrow \frac{2}{x} = \frac{1}{\sqrt{x^{2} - 160^{2}}}

\begin{aligned} x = 2 x - 160 \end{aligned}

x = 320 / v 3

\begin{array}{ll} GA + GB + GC = 120 + 160 \sqrt{3} \end{array}

= 40 (4 \sqrt{3} + 3)

Maxima and Minima Exercise Case Study Based Questions Question 8(i)

Answer: $x^{2} + y^{2} = 400$
Hint: use pythagoras theorem.
Solution:
Length

= 2 x

Breadth

= 2 y

\begin{aligned} {AB}^{2} + {BC}^{2} = {AC}^{2} \end{aligned}

(2 x)^{2} + (2 y)^{2} = (40)^{2}

4 x^{2} + 4 y^{2} = 1600

x^{2} + y^{2} = 400

Maxima and Minima Exercise Case Study Based Questions Question 8(ii)

Answer: $A = 4 x \sqrt{400 - x^{2}}$
Hint: use the formula of area of rectangle.
Solution:
Area

= l \times b

A = 2 x \times 2 y

As we know,

\begin{aligned} x^{2} + y^{2} = 400 \end{aligned}

y^{2} = 400 - x^{2}

y = \sqrt{400 - x^{2}}

A = 2 x \times 2 \sqrt{400 - x^{2}}

A = 4 x \sqrt{400 - x^{2}}

Maxima and Minima Exercise Case Study Based Questions Question 8(iii)

Answer: $\frac{8 (200 - x^{2})}{\sqrt{400 - x^{2}}}$
Hint: use the concept of maxima and minima.
Solution:

\begin{aligned} x^{2} + y^{2} & = 400 \end{aligned}

y^{2} = 400 - x^{2}

A = 1 \times b

A = 2 x \times 2 \sqrt{400 - x^{2}}

= 4 x \sqrt{400 - x^{2}}

\begin{aligned} \frac{d A}{d x} = 4 \sqrt{400 - 4 x^{2}} + 4 (\frac{(- 2 x^{2})}{2 \sqrt{400 - x^{2}}}) \end{aligned}

\frac{d A}{d x} = \frac{8 (200 - x^{2})}{\sqrt{40 x - x^{2}}}

Maxima and Minima Exercise Case Study Based Questions Question 8(iv)

Answer:

800 m^{2}

Hint: use the concept of maxima and minima.
Solution:
As we know,

\begin{aligned} \frac{dA}{dx} = 0 \end{aligned}

\frac{dA}{dx} = \frac{8 (200 + x^{2})}{\sqrt{400 - x^{2}}}

\frac{d A}{dx} = 0

f (200 + x^{2}) = 0

1600 + 8 x^{2} = 0

\begin{aligned} 8 x^{2} = - 1600 \end{aligned}

x^{2} = - 1600

x = \pm 400

x = 400 for x

for, 2 x = 2 \times 400 = 800 m^{2}

Thus is the maximum area of the garden.

Maxima and Minima Exercise Case Study Based Questions Question 8(v)

Answer:

400 (π - 2) m^{2}

Hint: use basic cgfy concept.
Solution:
Other part is converted into circle

= x (π - 2)

= 400 (π - 2) m^{2}

The Sharma class 12th exercise CSBQ is essential for the students who face difficulties in learning the mathematical concepts. A teacher or a tutor cannot assist a student any time whenever they have doubts. Having a copy of the RD Sharma Class 12th exercise CSBQ book gives the students a clarity regarding their doubts. Therefore, no one would feel perplexed while doing their homework or assignments.

The chapter 17 in the class 12 mathematics syllabus has five exercises in total. The CSBQ part consists of 40 questions to be answered by the students. It is obvious that the teachers will not help them with finding the answers for all these questions. The homework can be completed by the students using the RD Sharma Class 12th Exercise CSBQ Solution.

Some of the topics that this portion covers are finding the maximum and minimum values of a function, finding the properties of maxima and minima, solving the higher-order derivative test, definition of maximum and minimum, point of inflection and point of inflexion. The RD Sharma Class 12 Solutions Chapter 17 Exercise CSBQ will be an effective reference material in solving the doubts regarding all these topics.

Benefits of using the RD Sharma Class 12th Exercise CSBQ in exam preparation:

All the solutions are given by an expert team who have contributed their knowledge for the welfare for the students. The revised set of solutions gives the assurance that they are accurate and have no mistakes in it.
The questions in the sample papers and the solutions present in the RD Sharma Class 12 Chapter 12 Exercise CSBQ are changed according to the updates made in the NCERT syllabus books.
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Frequently Asked Questions (FAQs)

1. Which reference book would a class 12 handling teacher recommend to her students?

Most of the teachers suggest the RD Sharma Class 12 Chapter 17 solutions book to their students.

2. Do the RD Sharma books follow the NCERT pattern?

The RD Sharma books are framed according to the NCERT pattern in a way that the CBSE students can gain knowledge from it.

3. What makes the RD Sharma books essential for a class 12 student?

Along with the solutions for all the questions given in the textbook, the RD Sharma books contain various sample papers and additional questions for the students to practice well before their exams. This makes it very useful for the class 12 students.

4. Which is the best online site that contains the RD Sharma reference books?

The Career 360 website contains the entire set of RD Sharma reference materials where everyone can access and download the books.

5. How many CSBQ from this chapter does the RD Sharma solution book covers?

There are 40 CSBQ in this chapter for which all the solutions are provided in the RD Sharma Class 12 Chapter 17 Solutions material.

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