Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.3 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.3

Question 1.

Determine the maximum and minimum values of the following functions:

(i) f(x) = 2x^{3} – 21x^{2} + 36x – 20

Solution:

f(x) = 2x^{3} – 21x^{2} + 36x – 20

∴ f'(x) = \(\frac{d}{d x}\)(2x^{3} – 21x^{2} + 36x – 20)

= 2 × 3x^{2} – 21 × 2x + 36 × 1 – 0

= 6x^{2} – 42x + 36

and f”(x) = \(\frac{d}{d x}\)(6x^{2} – 42x + 36)

= 6 × 2x – 42 × 1 + 0

= 12x – 42

f'(x) = 0 gives 6x^{2} – 42x + 36 = 0.

∴ x^{2} – 7x + 6 = 0

∴ (x – 1)(x – 6) = 0

∴ the roots of f'(x) = 0 are x_{1} = 1 and x_{2} = 6.

For x = 1, 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)^{3} – 21(1)^{2} + 36(1) – 20

= 2 – 21 + 36 – 20

= -3

For x = 6, 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)^{3} – 21(6)^{2} + 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.

(ii) f(x) = x . log x

Solution:

f(x) = x . log x

f'(x) = \(\frac{d}{d x}\)(x.log x)

= x.\(\frac{d}{d x}\)(log x) + log x.\(\frac{d}{d x}\)(x)

= x × \(\frac{1}{x}\) + (logx) × 1

= 1 + log x

and f”(x) = \(\frac{d}{d x}\)(1 + logx)

= 0 + \(\frac{1}{x}\)

= \(\frac{1}{x}\)

Now, f'(x) = 0, if 1 + log x = 0

i.e. if log x = -1 = -log e

i.e. if log x = log(e^{-1}) = log \(\frac{1}{e}\)

i.e. if x = \(\frac{1}{e}\)

When x = \(\frac{1}{e}\), f”(x) = \(\frac{1}{(1 / e)}\) = e > 0

∴ by the second derivative test,

f is minimum at x = \(\frac{1}{e}\)

Minimum value of f at x = \(\frac{1}{e}\)

= \(\frac{1}{e}\) log(\(\frac{1}{e}\))

= \(\frac{1}{e}\) log(e^{-1})

= \(\frac{1}{e}\) (-1) log e

= \(\frac{-1}{e}\) ……..[∵ log e = 1]

Hence, the function f has minimum at x = \(\frac{1}{e}\) and minimum value is \(\frac{-1}{e}\).

(iii) f(x) = x^{2} + \(\frac{16}{x}\)

Solution:

f'(x) = 0 gives 2x – \(\frac{16}{x^{2}}\) = 0

∴ 2x^{3} – 16 = 0

∴ x^{3} = 8

∴ x = 2

For x = 2, f”(2) = 2 + \(\frac{32}{(2)^{3}}\) = 6 > 0

∴ by the second derivative test, f has minimum at x = 2 and minimum value of f at x = 2

f(2) = (2)^{2} + \(\frac{16}{2}\)

= 4 + 8

= 12

Hence, the function f has a minimum at x = 2 and a minimum value is 12.

Question 2.

Divide the number 20 into two parts such that their product is maximum.

Solution:

Let the first part of 20 be x.

Then the second part is 20 – x.

∴ their product = x(20 – x) = 20x – x^{2} = f(x) …..(Say)

∴ f'(x) = \(\frac{d}{d x}\)(20x – x^{2})

= 20 × 1 – 2x

= 20 – 2x

and f”(x) = \(\frac{d}{d x}\)(20 – 2x)

= 0 – 2 × 1

= -2

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

i.e. 20 – 2x = 0 is x = 10

and f”(10) = -2 < 0

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

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

Question 3.

A metal wire of 36 cm long is bent to form a rectangle. Find its dimensions where its area is maximum.

Solution:

Let x cm and y cm 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)

Let f(x) = x(18 – x) = 18x – x^{2}

Then f'(x) = \(\frac{d}{d x}\)(18x – x^{2})

= 18 × 1 – 2x

= 18 – 2x

and f”(x) = \(\frac{d}{d x}\)(18 – 2x)

= 0 – 2 × 1

= -2

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

Hence, the rectangle is a square of side 9 cm.

Question 4.

The total cost of producing x units is ₹(x^{2} + 60x + 50) and the price is ₹(180 – x) per unit. For what units is the profit maximum?

Solution:

Let the number of units sold be x.

Then profit = S.P. – C.P.

∴ P(x) = (180 – x)x – (x^{2} + 60x + 50)

∴ P(x) = 180x – x^{2} – x^{2} – 60x – 50

∴ P(x) = 120x – 2x^{2} – 50

P'(x) = \(\frac{d}{d x}\)(120x – 2x^{2} – 50)

= 120 × 1 – 2 × 2x – 0

= 120 – 4x

and P”(x) = \(\frac{d}{d x}\)(120 – 4x)

= 0 – 4 × 1

= -4

P'(x) = 0 if 120 – 4x = 0

i.e. if x = 30 and P”(30) = -4 < 0

∴ by the second derivative test, P(x) is maximum when x = 30.

Hence, the number of units sold for maximum profit is 30.