OP Malhotra Class 12 Maths Solutions Chapter 26 Application of Calculus in Commerce and Economics Ex 26(c)

Continuous practice using S Chand Class 12 Maths Solutions Chapter 26 Application of Calculus in Commerce and Economics Ex 26(c) can lead to a stronger grasp of mathematical concepts.

S Chand Class 12 ICSE Maths Solutions Chapter 26 Application of Calculus in Commerce and Economics Ex 26(c)

Question 1.
The demand for a certain product is represented by the equation p = 20 + 5x – 3x² where x is the number of units demanded and;? is the price per unit. Find
(i) Total revenue
(ii) The marginal revenue
(iii) The marginal revenue when 2 units are sold.
Solution:
Given p = 20 + 5x – 3x²
(i) ∴ Total revenue function R(x) = p × x = 20x + 5x² – 3x³
(ii) Marginal revenue function (MR) = \(\frac { d }{ dx }\) R(x) = 20 + 10x – 9x²
(iii) (MR)_x=2= 2 0 + 10 ×2 – 9 × 2² = 40 – 36 = 4

Question 2.
(i) The demand function of a monopolist is given by p = 100 – x – x². Find (a) the revenue function, (b) marginal revenue function.
(ii) If \(p=\frac{150}{q-5}-6 q\) represents the demand function for a product, where p is the price per unit for q units. Determine the marginal revenue function.
(iii) If \(p=\frac{5}{q-5}-6 q\) represents the demand function for a product, where p is the price per unit of q units, find the marginal revenue.
Solution:
(i) Given demand function p = 100 – x – x²
(a) ∴ revenue function R(x) = px = 100x – x² – x³
(b) Marginal revenue function MR = \(\frac { d }{ dx }\) R(x) = 100 – 2x – 3x²

(ii) Given \(p=\frac{150}{q^2+2}-4\)

Question 3.
The total revinue received from the sale of x units of a product is given by R(x) = 36x + 3x² + 5 Find .
(i) the average revenue
(ii) the marginal revenue
(iii) the marginaXand average revenue when x = 5
(iv) the actual revenue from selling 50 th item.
Solution:
Given total revenue function R(x) = 36x + 3x² + 5
(i) ∴ average revenue function (AR) = \(\frac{\mathrm{R}(x)}{x}\) = 36 + 3x + \(\frac { 5 }{ x }\)
(ii) Marginal revenue function (MR) = \(\frac { d }{ dx }\) R(x) = 36 + 6x
(iii) (AR)_{x = 5} = 36 + 15 + \(\frac { 5 }{ 5 }\) = 36 + 15 + 1 = 52 and MR_{x = 5} = 36 + 6 × 5 = 66
(iv) Required actual revenue from selling 50 th item = R(50) – R(49)
= 36 × 50 + 3 × 50² + 5 – 36 × 49 – 3 × 49² – 5
= 36 × (50 – 49) + 3 × (50 – 49)(50 + 49)
= 36 + 297 = 333

Question 4.
The demand function for a monopolist is given by x = 100 – 4 p. Find
(i) total revenue function
(ii) average revenue function
(iii) marginal revenue function
(iv) price and quantity at which MR =0
Solution:
Given demand function x = 100 – 4p ⇒ 4p = 100 – x ⇒ p = 25 – \(\frac { x }{ 4 }\)
(i) ∴ Total revenue function = p × x = 25x – \(\frac{x^2}{4}\) = R(x)
(ii) average revenue function = \(\frac{\mathrm{R}(x)}{x}\) = 25 – \(\frac{x}{4}\)
(iii) Marginal revenue function = MR = \(\frac { d }{ dx }\) R(x) = 25 – \(\frac { 2x }{ 4 }\) = 25 – \(\frac{x}{2}\)
(iv) When MR = 0 ⇒ 25 – \(\frac{x}{2}\) = 0 ⇒ x = 50 units
∴ required price p = 25 – \(\frac{x}{4}\) at x = 50; p = 25 – \(\frac{50}{4}\) = \(\frac{25}{2}\) = ₹ 12.5

Question 5.
A monopolists demand function for one of its products is p(x) = ax + b. He knows that $b$ can sell 1400 units when the price is ₹ 4 per unit and he can sell 1800 units at a price of ₹? per unit. Find the total average and marginal revenue functions. Also find the price unit when the marginal revenue is zero.
Solution:
Given demand function p(x) = ax + b …(1)
given x = 1400 when p = ₹ 4 ∴ from (1); 4 = 1400 a + b …(2)
When x = 1800; p = ₹ 2 ∴ from (1); 2 = 1800 a + b ….(3)
eqn. (3) – eqn. (2); we have
-2 = 400 a ⇒ a = –\(\frac { 1 }{ 200 }\)
∴ from (2); 4 = -7 + b ⇒ b = 11
∴ from (1); p(x) = –\(\frac { x }{ 200 }\) + 11
∴ revenue function = R(x) = p(x) × x = –\(\frac{x^2}{200}\)
Thus Average revenue function AR = \(\frac{\mathrm{R}(x)}{x}\) = – \(\frac{x}{200}\) + 11
Marginal revenue function MR = \(\frac{d}{dx}\)R(x) = \(\frac{-2 x}{200}\) + 11 = \(\frac{-x}{100}\) + 11
When MR = 0 ⇒ \(\frac{-x}{100}\) + 11 = 0 ⇒ x = 1100
∴ from (1); p = \(\frac{-1100}{200}\) + 11 = ₹ 5.50

Question 6.
Suppose the consumers will demand 40 units of a product when the price is ₹ 12 per unit and 25 units when the price is ₹ 18 each. Find the demand function, assuming that is linear. Also, determine the total revenue function, the average revenue function and the marginal revenue function.
Solution:
Let the demand function be x = ap + b …(1)
Where p is the price per unit when x units are demanded
given when x = 40; p = 12 ∴ from (1); 40 = 12a + b ….(2)
when x = 25; p = 18 ∴ from (1); 25 = 18a + b …(3)
eqn. (3) – eqn. (2) gives; – 15 = 6 a ⇒ a = \(\frac{-15}{6}\) = \(\frac{-15}{6}\)
∴ from (3); 25 = – 45 + b ⇒ b = 70
∴ from (1); x = \(\frac{-5}{2}\) p + 70
∴ p = (70 – x) \(\frac{2}{5}\) = 28 – \(\frac{2}{5}\)x
Thus total revenue function = R(x) = px = 28x – \(\frac{2}{5}\)x²
average revenue function = AR = \(\frac{\mathrm{R}(x)}{x}\) = 28 – \(\frac{2}{5}\)x
and Marginal revenue function (MR) = \(\frac{d}{dx}\)R(x) = 28 – \(\frac{4}{5}\)x

Question 7.
A firm knows that the demand function for one of its products is linear. It also knows that it can sell 1000 units when the price is ₹ 4 per unit, and it can sell 1500 units when the price is ₹ 2 a unit. Determine
(i) the demand function,
(ii) the total revenue function,
(iii) the average revenue function,
(iv) the marginal revenue function.
Solution:
(i) Let the required demand function be
x =a p + b ..(1)
where p is the price per unit when x units are demanded.
When x=1000 ; p = 4 ∴ from (1); 1000 =4a + b
When x = 1500; p = 2 ∴ from (1); 1500 = 2a + b
eqn. (3) – eqn. (2) gives ; 500 = – 2a ⇒ a = – 250
∴ from (2); 1000 = -1000 + b ⇒ b = 2000
Thus eqn. (1) becomes; x = – 250p + 2000
∴ 250p = 2000 – x ⇒ p = \(\frac{2000}{250}\) – \(\frac{x}{250}\) ⇒ p = 8 – \(\frac{x}{250}\)
(ii) Therefore total revenue function R(x) = p × x = 8x – \(\frac{x^2}{250}\)
(iii) Average revenue function AR = \(\frac{\mathrm{R}(x)}{x}\) = 8 – \(\frac{x}{250}\)
(iv) Marginal revenue function MR = \(\frac{d}{d x}\)R(x) = 8 – \(\frac{x}{125}\)

Question 8.
Find the relationship between the slopes of marginal revenue curve and the average revenue curve, for the demand function p =a – bx.
Solution:
Given demand function p = a – bx
∴ Revenue function = R(x) = p × x = ax – bx²
∴ Average revenue function AR = \(\frac{R(x)}{x}\) = a – bx
Thus slope of AR curve = \(\frac{d}{d x}\) (AR) = -b
Marginal revenue function MR = \(\frac{d}{d x}\)R(x) = a – 2bx
∴ slope of MR curve = \(\frac{d}{d x}\) MR = -2b
Thus slope of MR curve is twice the slope of AR curve.

Question 9.
In the production unit of a firm it is found that the total number of unites produced is dependent upon the number of workers and is obtained by the relation x = 25 n(n³ + 36)^-1/2. The demand function of the product is p =\(\frac{250}{x+15}\). Determine the marginal revenue when n = 4.
Solution:
Given demand function p = \(\frac{250}{x+15}\)
∴ Total revenue function = R(x) = p × x = \(\frac{250}{x+15}\)
∴ Marginal revenue function MR = \(\frac{d}{d x}\)R(x) = 250 \(\left[\frac{x+15-x}{(x+15)^2}\right]\) = \(\frac{250 \times 15}{(x+15)^2}\) …(1)
given x = 25n (n³ + 36)^-1/2
When n = 4 ∴ x = 25 × 4 (64 + 36)^-1/2 = 100 × (100)^-1/2
⇒ x = \(\frac{100}{\sqrt{100}}\) = \(\frac{100}{10}\) = 10
When x = 10 units ∴ from (1); we have MR = \(\frac{250 \times 15}{(25)^2}\) = \(\frac{150}{25}\) = 6

Question 10.
For the demand function \(p=\frac{a}{x+b}-c\), where ab > 0, show that the narginal revenue decreases with the increase of x.
Solution:
Given demand function \(p=\frac{a}{x+b}-c\)
∴ revenue function = R(x) = px = \(\left[\frac{a}{x+b}-c\right] x\)
Thus marginal revenue function (MR) = \(\frac{d}{dx}\)R(x) = a\(\left[\frac{x+b-x}{(x+b)^2}\right]\) – c = \(\frac{a b}{(x+b)^2}\) – c
∴ \(\frac{d}{dx}\)(MR) = \(\frac{d}{dx}\)\(\left[\frac{a b}{(x+b)^2}-c\right]\) = – \(\frac{2 a b}{(x+b)^3}\) < 0 ∀ x > 0 [∵ ab > 0]
Thus, marginal revenue decreases with the increase of x.