RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5 (Updated For 2024)

RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5: The RD Sharma book solution uses a structured approach to better understand and understand concepts. You can easily download the solution from the provided PDF. These solutions are designed by a group of experts in the field to eliminate students’ doubts. The maximum and minimum values ​​of functions in your domain are the main topics covered in this exercise. This exercise consists of two levels, in order of increasing difficulty.

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Access RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5

1. Determine two positive numbers whose sum is 15 and the sum of whose squares is minimum.

Solution:

Which implies S is minimum when a = 15/2 and b = 15/2.

2. Divide 64 into two parts such that the sum of the cubes of two parts is minimum.

Solution:

Let the two positive numbers be a and b.

Given a + b = 64 … (1)

We have, a³ + b³ is minima

Assume, S = a³ + b³

(From equation 1)

S = a³ + (64 – a)³

of S

Hence, the two number will be 32 and 32.

3. How should we choose two numbers, each greater than or equal to –2, whose sum is ½ so that the sum of the first and the cube of the second is minimum?

Solution:

dS/da = 0

1 + 3(½ – a)² (-1) = 0

1 – 3(½ – a)² = 0

4. Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.

Solution:

Let the given two numbers be x and y. Then,

x + y = 15 ….. (1)

y = (15 – x)

Now we have z = x² y³

z = x² (15 – x)³ (from equation 1)

5. Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm³, which has the minimum surface area?

Solution:

Let r and h be the radius and height of the cylinder, respectively. Then,

Volume (V) of the cylinder = πr² h

⟹ 100 = πr² h

⟹ h = 100/ πr²

Surface area (S) of the cylinder = 2 πr² + 2 πr h = 2 πr² + 2 πr × 100/ πr²

6. A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by

Find the point at which M is maximum in each case.

Solution:

7. A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the circle and the square is minimum?

Solution:

Suppose the given wire, which is to be made into a square and a circle, is cut into two pieces of length x and y m respectively. Then,

x + y = 28 ⇒ y = (28 – x)

We know that the perimeter of a square, 4 (side) = x

Side = x/4

Area of square = (x/4)² = x²/16

Circumference of a circle, 2 π r = y

r = y/ 2 π

8. A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the wire should be cut so that the sum of the areas of the square and triangle is minimum?

Solution:

Suppose the wire, which is to be made into a square and a triangle, is cut into two pieces of length x and y respectively. Then,

x + y = 20 ⇒ y = (20 – x) …… (1)

We know that the perimeter of a square, 4 (side) = x

Side = x/4

Area of square = (x/4)² = x²/16

Again we know that the perimeter of a triangle, 3 (side) = y.

Side = y/3

Hence, the wire of length 20 m should be cut into two pieces of lengths

9. Given the sum of the perimeters of a square and a circle, show that the sum of their areas is least when one side of the square is equal to diameter of the circle.

Solution:

Let us say the sum of the perimeter of the square and circumference of the circle be L

Given the sum of the perimeters of a square and a circle.

Assuming the side of the square = a and the radius of the circle = r

Then, L = 4a + 2πr ⇒ a = (L – 2πr)/4… (1)

Let the sum of the area of the square and circle be S

So, S = a² + πr²

10. Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.

Solution:

11. Two sides of a triangle have lengths ‘a’ and ‘b’ and the angle between them is θ. What value of θ will maximize the area of the triangle? Find the maximum area of the triangle also.

Solution:

It is given that two sides of a triangle have lengths a and b, and the angle between them is θ.

Let the area of the triangle be A

12. A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Also, find this maximum volume

Solution:

Given side length of big square is 18 cm

Let the side length of each small square be a.

If by cutting a square from each corner and folding up the flaps we will get a cuboidal box with

Length, L = 18 – 2a

Breadth, B = 18 – 2a and

Height, H = a

Assuming, volume of box, V = LBH = a (18 – 2a)²

Condition for maxima and minima is

13. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off squares from each corners and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?

Solution:

Given the length of the rectangle sheet = 45 cm

Breath of rectangle sheet = 24 cm

Let the side length of each small square be a.

If by cutting a square from each corner and folding up the flaps we will get a cuboidal box with

Length, L = 45 – 2a

Breadth, B = 24 – 2a and

Height, H = a

Assuming, volume of box, V = LBH = (45 – 2a)(24 – 2a)(a)

Condition for maxima and minima is

(45 – 2a) (24 – 2a) + (- 2) (24 – 2a) (a) + (45 – 2a) (- 2)(a) = 0

4a² – 138a + 1080 + 4a² – 48a + 4a² – 90a = 0

12a² – 276a + 1080= 0

a² – 23a + 90= 0

a = 5, 18

RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5: Important Topics From The Chapter

Some of the crucial topics of  RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5 are enlisted here.

• Maximum and minimum values of a function in its domain
• Definition and meaning of maximum
• Definition and meaning of minimum
• Local maxima
• Local minima
• First derivative test for local maxima and minima
• Higher-order derivative test
• Theorem and algorithm based on higher derivative test
• Point of inflection
• Point of inflection
• Properties of maxima and minima
• Maximum and minimum values in a closed interval
• Applied problems on maxima and minima

We have included all the information regarding CBSE RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5. If you have any query feel free to ask in the comment section. 

FAQ: RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5