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Edited By Satyajeet Kumar | Updated on Jan 21, 2022 02:40 PM IST

RD Sharma class 12th exercise 17.5 is evidently one of the most trusted and beneficial books for students in high school. It covers comprehensive knowledge on the subject and imparts accurate information on every concept. This guidance aids students to gain clarity on complex concepts and prepares them to score high in their exams.

Experts were in charge of preparing the RD Sharma solutions and their knowledge reflects on the quality of answers and the techniques that are used in the making of RD Sharma Class 12 solution of Maxima and minima Exercise 17.5. These solutions will help prepare you for your recent board examinations and provide you with plenty of practice to increase confidence. With the help of RD Sharma Class 12th exercise 17.5, students can also check their homework and complete them with ease.

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This Story also Contains

- RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima - Other Exercise
- Maxima and Minima Excercise:17.5
- Maxima and minima exercise 17.5 question 3
- Maxima and minima exercise 17.5 question 4
- Maxima and minima exercise 17 5 question 5
- Maxima and minima exercise 17.5 question 6 sub question (i)
- Maxima and minima exercise 17.5 question 6 sub question (ii)
- Maxima and Minima exercise 17.5 question 7.
- Maxima and Minima exercise 17.5 question 9.
- Maxima and Minima exercise 17.5 question 12.
- Maxima and Minima exercise 17.5 question 13.
- Answer:
- Maxima and Minima exercise 17.5 question 14
- Maxima and Minima exercise 17.5 question 15.
- Maxima and Minima exercise 17.5 question 16.
- Maxima and Minima exercise 17.5 question 19.
- Answer:
- Maxima and Minima exercise 17.5 question 21 maths.
- Answer:
- Maxima and Minima exercise 17.5 question 23
- Maxima and Minima exercise 17.5 question 25
- Maxima and Minima exercise 17.5 question 26
- Maxima and Minima exercise 17.5 question 28
- maxima and minima exercise 17.5 question 29
- maxima and minima exercise 17.5 question 30
- RD Sharma Chapter-wise Solutions

- Chapter 17 - Maxima and Minima - Ex 17.1
- Chapter 17 - Maxima and Minima - Ex 17.2
- Chapter 17 - Maxima and Minima - Ex 17.3
- Chapter 17 - Maxima and Minima - Ex 17.4
- Chapter 17 - Maxima and Minima - Ex FBQ
- Chapter 17 - Maxima and Minima - Ex CSBQ
- Chapter 17 - Maxima and Minima - Ex MCQ
- Chapter 17 - Maxima and Minima- Ex VSA

Maxima and minima exercise 17.5 question 1

Answer:

**Hint:** For maximum or minimum value of z must have **Given:****Solution:** ..........(1)

Now,

(from equation 1)

For maximum or minimum value of z

Substituting in (1)

z is minimum when x = y =

x= 32,33**Hint: **For maximum or minimum value of z must have **Given:** 64 divide into two parts, sum of cubes of two parts is minimum**Solution:** Suppose 64 divide in two parts x and 64 - x then,

For maximum and minimum value of z

z is minimum when 64 is divided into two parts 32 and 32

and .....(1)

For maximum or minimum value of z

z is minimum when

At

z is minimum when

Substituting the value in (1)

So the two required numbers are

Maxima and minima exercise 17.5 question 4

Answer:

x + y = 15 .......(1)

Now,

from (1)

For minimum and maximum value of z

At x =6

Thus z is maximum when x =6 and y =9

x + y = 15 .......(1)

Now,

from (1)

For minimum and maximum value of z

At x =6

Thus z is maximum when x =6 and y =9

Volume (V) of cylinder =

Surface area (S) of the cylinder =

For maximum or minimum,

Thus surface area is minimum when

Cylinder with radius has min surface area.

For maximum or minimum ,

So m is maximum at

For maximum or minimum,

So m is minimum at

Perimeter of square 4a=x

Area of square

Circumference of circle

Area of circle

z = Area of square + Area of circle

For maximum or minimum value of z

from equation 1

Thus z is maximum when and

Perimeter of square = 4a = x

Area of square

Perimeter of triangle 3a=y

Area of triangle

Now,

z =Area of square + area of triangle

For minimum or maximum of z

from equation (1)

Thus z is maximum when x

.......(1)

Now,

.....rom equation (1)

So the sum of the areas, A is least when

From (1) and (2) we get

Side of square = Diameter of circle

Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.

Maxima and Minima exercise 17.5 question 10

Let the base of the triangle be x, and height be y

As area of triangle A,

Also

So, is point of maxima

Largest possible area of triangle,

square units

Maxima and Minima exercise 17.5 question 11.

For maxima and minima

Also or

i.e,

is teh point of maxima

The maximum area of triangle

Let the square side be cutoff be xcm. Then, the length and breadth of the box will be (18-2x)cm each and height of the box will be xcm

Volume of the box

For maxima or minima V'(x) =0

x=9 or x =3

if x=9, then l and b will become 0

So,

if

Now,

x= 3 is the point of maximum

Then, the l and b, h of the loon will be (45-x), (25-x),xcm

Volume of the box V = (45-x) (25-x)x

Divide by 12

Thus volume of the box is max when x = 5cm.

Let l,b and h be the length, breadth and height of the tank

h = 2cm

Volume of tank = 8m

Volume of the tank =

Area of base

Area of 4 walls,

The length cannot be negative.

now

Thus the area is minimum when l=2,

Cost of the building the base =

Cost of the building walls

Total cost = Rs (280 +720) = 1000

Let the dimensions of the rectangular part be x,y

Radius of the semicircle

Total perimeter = 10

(1)

Area

..... from (1)

Substitute value of x in eqn (1)

Thus the area is max when x

So,

Let the dimensions of the rectangular part x, y

Perimeter of the window ⇒

.....(1)

Area =

Substitute x value in eqn (1)

Thus the area is max when

Maxima and Minima exercise 17.5 question 17.

Answer:

.....(1)

Volume of the cylinder V

Square on both the sides

..........from (1)

Now,

Subsitute r value in eqn (1)

Volume of the cylinder is max when,

Maxima and Minima exercise 17.5 question 18

Answer:

......(1)

Area = xy

Square on both sides

......from (1)

Now,

Substitute y value in eqn (1)

Thus the area is maximum when and

Area = xy

Now,

For minima

Substitute V value in equation (1)

Maxima and Minima exercise 17.5 question 20.

Answer:

.......(1)

Volume of cone

..... from (1)

Now,

Substituting y value in eqn (1)

Volume is maximum when

(Diameter of sphere)

Hence proved.

Slant height,

Curved surface area,

So,

Since, and for

Curved surface area for or

Maxima and Minima exercise 17.5 question 22.

Answer:

Let ABC be an isoceles triangle inscribed in the circle with the radius a such that

AB=AC

Area of triangle ACA

Can be inscribed in

Let ABC an isoceles triangle with AB = AC =x and BC =y and a circle with centre O and radius r inscribed in triangle ABC.

AO= 2r, OF =r

Using phythagoras theorem in triabgle ABF

..........(1)

Again for ,

BD=BF, EC=FC

Now, AD + DB = x

........(2)

From (2)

Perimeter = 2x +y

Perimeter

Maxima and Minima exercise 17.5 question 24

.....(1)

..... from (1)

Now,

Substitute the value of b in (1)

So, the volume is maximum when a = 12cm and b = 6cm

Now,

....... from (1)

Now,

So, the volume is maximum when

Volume

.........(1)

Surface area =

......from (1)

Now,

So, the surface area is maximum when r = 7

.......(1)

Now,

......... from (1)

Volume of maximum

Maximum volume

.........(1)

Now,

......from (1)

Squaring on the both sides

Substituting the value of x in equation (1)

So z = x + y is maximum when

.........(1)

Also,

using the distance formula

Now,

....... from (1)

Substituting the value of x is equal (1)

y =3

The required nearest point is

........(1)

Also,

using distance formula

Now,

...... from (1)

Substituting the value of y in equation (1)

The required nearest point is (4,-4)

maxima and minima exercise 17.5 question 31

Answer:

....... (1)

Also,

using distance formula

Now,

..... from (1)\

Substituting the value of y in eqn (1)

y =2

The required nearest point is (4,2)

maxima and minima exercise 17.5 question 32

Answer:

**Hint:** For maximum or minimum value of z must have **Given:** Let the point (x,y) on the curve nearest (0,5)

.........(1)

Also,

using the distance formula**Solution:**

Now,

....... from (1)

Substituting the value of x in eqn (1)

y =4

The required nearest point .

maxima and minima exercise 17.5 question 33

Answer:

......(1)

The required nearest point is

maxima and minima exercise 17.5 question 34

Answer:

(2, 2)........(1)

using the distance formula

Now,

........ from (1)

Substituting the value of y in eqn (1)

x = 2

The required nearest point is (2,2)

maxima and minima exercise 17.5 question 35 maths textbook solution.

Answer:

......(1)

Slope =

Now,

x = 1

Substituting the value of x in eqn(1)

So the slope is maximum x= 1, y = -23 at (1,-23)

maximum slope

Maxima and Minima exercise 17.5 question 36.

Answer:

Now,

Profit is the maximum if daily output is 10 units.

Maxima and Minima exercise 17.5 question 37.

Answer:

240 itemsNow,

Profit is maximum if 240 items are sold.

Maxima and Minima exercise 17.5 question 38.

Answer:

Since volume, v is constant →i

Let S denote the total surface area, then we have

→fromi

…(ii)

Differentiating S w.r.t 'l', we get

…(iii)

For maximum or minimum of S, we have

[using (iii)]⇒]

…(iv)

Again, differentiating S w.r.t 'l', then [using (iii)]

Hence the surface area is minimum, Substitute the value of from (iv) in eqni, we get

Hence proved.

Maxima and Minima exercise 17.5 question 39.

Answer:

Let l,b,h be the length , breadth and height of the box

Volume of the box = c

l = 2b ....(1)

c = lbh

.......(2)

Let the cost of material required to the bottom be .

Cost of material required for 4 wall stand top =

Total cost ,

[From (1) and (2)}

cost of the minimum when

Sub b in eqn(1) , eqn (2)

Maxima and Minima exercise 17.5 question 40.

Answer:

Let r be the radius of the sphere, x is side of the cube and S be the sum of the surface area of both of them

......(1)

Sum of Volumes,

From eqn (1)

.........(2)

from eqn (1),

Volume is minimum when x = 2r

Maxima and Minima exercise 17.5 question 41

Answer:

.........(1)

Total surface area =

.....from (1)

Hence proved

Maxima and Minima exercise 17.5 question 42.

Answer:

Let the breadth, height, strength of the beam be b,h,S

.......(1)

Stength beam,

......from (1)

Sub b value in (1)

So the strength of the beam is maximum when

Maxima and Minima exercise 17.5 question 43

Answer:

The equation of line passing through (1,4) with slope m is given by

........(1)

Sub (y=0) value in eqn (1)

Sub (x=0) value in eqn (1)

So the intercepts coordinates axes are

Sum is minimum at m = 2

So the sum is maximum at m = -2

Thus,

Maxima and Minima exercise 17.5 question 44

Answer:

Let x and y be the length and breadth of the rectangle

Area of page = 150

xy = 150

......(1)

Area of printed matter =

Sub x=15 value in eqn (1)

y = 10

Now,

Thus area of printed matter is max when x = 15 and y = 10

Maxima and Minima exercise 17.5 question 45.

Answer:

Acceleration,

Now,

So, acceleration is minimum at t =2

At =2, a = -260

Maxima and Minima exercise 17.5 question 46

Answer:

t = 2**Hint:** For maxima and minima value of V, we must have **Given:****Solution:**

On solving the equation we get

Now,

So, the velocity is maximum at

Again,

For the maximum or minimum value of a

Now,

So acceleration is minimum at t = 2

RD Sharma Class 12th exercise 17.5 comprises 37 level 1 questions and 9 level 2 questions including its subparts, and it incorporates elaborate solutions for all our queries and doubts. It includes the following concepts

- Maximum and minimum values of a function in its domain
- Local maxima and Local minima
- Point of inflection
- Maximum and minimum values in a closed interval
- Applied problems on maxima and minima

- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable

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