**Chapter 2 Applications of Derivatives Ex 2.4**

## Chapter 2 Applications of Derivatives Ex 2.4

**Question 1.Test whether the following functions are increasing or decreasing.**

**Question 2.Find the values of x for which the following functions are strictly increasing:**

**Question 3.Find the values of x for which the following functions are strictly decreasiong:**

**Question 4.Find the values of x for which the function f(x) = x ^{3} – 12x^{2} – 144x + 13(a) increasing(b) decreasing.Solution:**

**Question 5.**

**Question 6.**

(a) strictly increasing

(b) strictly decreasing.

**Solution:**

**Question 8.Show that f(x) = x – cos x is increasing for all x.Solution:**

f(x) = x – cos x

∴ f'(x) = d/dx (x – cos x)

= 1 – (-sin x)

= 1 + sin x

Now, -1 ≤ sin x ≤ 1 for all x ∈ R

∴ -1 + 1 ≤ 1 + sin x ≤ 1 for all x ∈ R

∴ 0 ≤ f'(x) ≤ 1 for all x ∈ R

∴ f'(x) ≥ 0 for all x ∈ R

∴ f is increasing for all x.

**Question 9.Find the maximum and minimum of the following functions:Solution:**

**Method 1 (Second Derivative Test):**

(a) f”(1) = 12(1) – 42 = -30 < 0

∴ by the second derivative test, f has maximum at x = 1

and maximum value of f at x = 1

f(1) = 2(1 – 21(1 + 36(1) – 20

= 2 – 21 + 36 – 20

= -3

(b) f”(6) = 12(6) – 42 = 30 > 0

∴ by the second derivative test, f has minimum at x = 6

and minimum value of f at x = 6

f(6) = 2(6 – 21(6 + 36(6) – 20

= 432 – 756 + 216 – 20

= -128.

Hence, the function f has maximum value -3 at x = 1 and minimum value -128 at x = 6.

Method 2 (First Derivative Test):

(a) f'(x) = 6(x – 1)(x – 6)

Consider x = 1

Let h be a small positive number. Then

f'(1 – h) = 6(1 – h – 1)(1 – h – 6)

= 6(-h)(-5 – h)

= 6h(5 + h)> 0

and f'(1 + h) = 6(1 + h – 1)(1 + h – 6)

= 6h(h – 5) < 0, as h is small positive number.

∴ by the first derivative test, f has maximum at x = 1 and maximum value of f at x = 1

f(1) = 2(1 – 21(1 + 36(1) – 20

= 2 – 21 + 36 – 20

= -3

(b) f'(x) = 6(x – 1)(x – 6)

Consider x = 6

Let h be a small positive number. Then

f'(6 – h) = 6(6 – h – 1)(6 – h – 6)

= 6(5 – h)(-h)

= -6h(5 – h) < 0, as h is small positive number

and f'(6 + h) = 6(6 + h – 1)(6 + h – 6) = 6(5 + h)(h) > 0

∴ by the first derivative test, f has minimum at x = 6

and minimum value of f at x = 6

f(6) = 2(6 – 21(6 + 36(6) – 20

= 432 – 756 + 216 – 20

= -128

Hence, the function f has maximum value -3 at x = 1

and minimum value -128 at x = 6.

(a) f”(2) = 6(2) – 18 = -6 < 0

∴ by the second derivative test, f has maximum at x = 2

and maximum value of f at x = 2

f(2) = (2 – 9(2 + 24(2)

= 8 – 36 + 48

= 20

**(v) f(x) = x log xSolution:**

Solution:

**Question 10.Divide the number 30 into two parts such that their product is maximum.Solution:**

Let the first part of 30 be x.

Then the second part is 30 – x.

= 0 – 2 × 1

= -2

The root of the equation f(x) = 0,

i.e. 30 – 2x = 0 is x = 15 and f”(15) = -2 < 0

∴ by the second derivative test, f is maximum at x = 15.

Hence, the required parts of 30 are 15 and 15.

**Question 11.Divide the number 20 into two parts such that the sum of their squares is minimum.Solution:**

Let the first part of 20 be x.

Then the second part is 20 – x.

= 4 × 1 – 0

= 4

The root of the equation f'(x) = 0,

i.e. 4x – 40 = 0 is x = 10 and f”(10) = 4 > 0

∴ by the second derivative test, f is minimum at x = 10.

Hence, the required parts of 20 are 10 and 10.

**Question 12.A wire of length 36 meters is bent in the form of a rectangle. Find its dimensions if the area of the rectangle is maximum.Solution:**

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

Then its perimeter is 2(x + y) = 36

x + y = 18

y = 18 – x

Area of the rectangle = xy = x (18 – x)

Now, f'(x) = 0, if 18 – 2x = 0

i.e. if x = 9

and f”(9) = -2 < 0

∴ by the second derivative test, f has maximum value at x = 9.

When x = 9, y = 18 – 9 = 9

∴ x = 9 cm, y = 9 cm

∴ the rectangle is a square of side 9 metres.

**Question 13.A ball is thrown in the air. Its height at any time t is given by h = 3 + 14t – . Find the maximum height it can reach.Solution:**

The height h at any t is given by h = 3 + 14t – 5

Hence, the maximum height the ball can reach = 12.8 units.

**Question 14.Find the largest size of a rectangle that can be inscribed in a semicircle of radius 1 unit, so that two vertices lie on the diameter.Solution:**

Let ABCD be the rectangle inscribed in a semicircle of radius 1 unit such that the vertices A and B lie on the diameter.

Let AB = DC = x and BC = AD = y.

Let O be the centre of the semicircle.

Join OC and OD. Then OC = OD = radius = 1.

Also, AD = BC and m∠A = m∠B = 90°.

∴ OA = OB

In right angled triangle OBC,

O + B = O

**Question 15.An open cylindrical tank whose base is a circle is to be constructed of metal sheet so as to contain a volume of π cu cm of water. Find the dimensions so that the quantity of the metal sheet required is minimum.Solution:**

Let x be the radius of the base, h be the height, V be the volume and S be the total surface area of the cylindrical tank.

Then V = π … (Given)

∴ πx

^{2}h = π

Now, S = 2πxh + π

∴ by the second derivative test, S is minimum when x = a

When x = a, from (1)

Hence, the quantity of metal sheet is minimum when radius height = a cm.

**Question 16.The perimeter of a triangle is 10 cm. If one of the sides is 4 cm. What are the other two sides of the triangle for its maximum area?Solution:**

Let ABC be the triangle such that the side BC = a = 4 cm.

Also, the perimeter of the triangle is 10 cm.

i.e. a + b + c = 10

∴ 2s = 10

∴ s = 5

Also, 4 + b + c = 10

∴ b + c = 6

∴ b = 6 – c

Let ∆ be the area of the triangle.

∴ by the second derivative test, ∆ is maximum when c = 3.

When c = 3, b = 6 – c = 6 – 3 = 3

Hence, the area of the triangle is maximum when the other two sides are 3 cm and 3 cm.

**Question 17.A box with a square base is to have an open top. The surface area of the box is 192 sq cm. What should be its dimensions in order that the volume is largest?Solution:**

Let x cm be the side of square base and h cm be its height.

Let V be the volume of the box.

∴ by the second derivative test, V is maximum at x = 8.

Hence, the volume of the box is largest, when the side of square base is 8 cm and its height is 4 cm.

**Question 18.The profit function P (x) of a firm, selling x items per day is given by P(x) = (150 – x)x – 1625. Find the number of items the firm should manufacture to get maximum profit. Find the maximum profit.Solution:**

Profit function P (x) is given by

Now, P'(x) = 0 gives, 150 – 2x = 0

∴ x = 75

and P”(75) = -2 < 0

∴ by the second derivative test, P(x) is maximum when x = 75

Maximum profit = P(75)

= (150 – 75)75 – 1625

= 75 × 75 – 1625

= 4000

Hence, the profit will be maximum, if the manufacturer manufactures 75 items and the maximum profit is 4000.

**Question 19.Find two numbers whose sum is 15 and when the square of one multiplied by the cube of the other is maximum.Solution:**

Let the two numbers be x and y.

Then x + y = 15

∴ y = 15 – x

Let P is the product of square of y and cube of x.

**Question 20.Show that among rectangles of given area, the square has least perimeter.Solution:**

Let x be the length and y be the breadth of the rectangle whose area is A sq units (which is given as constant).

Then xy = A

∴ y = A/x………(1)

Let P be the perimeter of the rectangle.

x = y

∴ rectangle is a square.

Hence, among rectangles of given area, the square has least perimeter.

**Question 21.Show that the height of a closed right circular cylinder of given volume and least surface area is equal to its diameter.Solution:**

Let x be the radius of base, h be the height and S be the surface area of the closed right circular cylinder whose volume is V which is given to be constant.

Hence, the surface area is least when height of the closed right circular cylinder is equal to its diameter.

**Question 22.Find the volume of the largest cylinder that can be inscribed in a sphere of radius ‘r’ cm.Solution:**

Let R be the radius and h be the height of the cylinder which is inscribed in a sphere of radius r cm.

Then from the figure,

**Question 23.**

Hence, the given function is increasing function on its domain.

**Question 24.**