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

Students can cross-reference their work with S Chand Class 12 Maths Solutions Chapter 26 Application of Calculus in Commerce and Economics Ex 26(e) to ensure accuracy.

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

Question 1.
(i) Determine the cost of 300 video cassette players, if the marginal cost (in rupees per unit) is given by MC(x) = \(\frac{x^2}{3}\) – 2x + 700.
(ii) Determine the cost of producing 1,000 T.V sets if the marginal cost (in rupees per unit) is given by C(x) = 0.003x² – 0.01x + 2.5.
(iii) Determine the cost of production of 3000 shirts if the marginal cost C in rupees per unit is C(x) = \(\frac { x }{ 100 }\) + 2.50
(iv) If the marginal cost is given by MC = 150 + e^0.5x, where x is the number of units of products in hundreds. If x increases from 2 to 4 , find the total increase in cost.
Solution:

Question 2.
(i) The marginal cost function of manufacturing x shoes is 6 + 10x – 6x². The total cost of producing a pair of shoes is ₹ 12. Find the total and average cost functions.
(ii) The marginal cost function of a firm is MC = 40 log x. Find the total cost function when the cost of producing one unit is ₹ 15.
(iii) The marginal cost function of a firm is MC = (log x)². Find the total cost function when the cost of producing one unit is ₹ 20.
Solution:
(i) Given Marginal cost function(MC) = 6 + 10x – 6x²
We know that MC = \(\frac{dC}{dx}\)

given when x = 2; C = 12
∴ from (1); 12 = 12 + 20 – 16 + K ⇒ K = -4
putting the value of K in eqn. (1); we have
C(x) = 6x + 5x² – 2x² – 4
∴ Average cost function = AC = \(\frac{\mathrm{C}(x)}{x}\) = 6 + 5x – 2x² – \(\frac{4}{x}\)

(ii) Given Marginal cost function (MC) = 40 log x
We Know that MC = \(\frac{d}{dx}\)(Cx)

Given when x = 1 ; C = ₹ 15
∴ from (1); we have
15 = 40[1 × 0 – 1] + K ⇒ K = 55
putting the value of K in eqn. (1); we get
C(x) = 40[x log x – x] + 55

(iii) Given marginal cost function MC = (log x)² ⇒ \(\frac{d}{dx}\)C(x) = (log x)²

Given when x =1 ; C = 20
∴ from (1); 20 = 0 + 0 + 2 + K ⇒ K = 18
putting the value of K in eqn. (1) ; we have
C(x) = x(log x)² – 2x logx + 2x + 18
be the required total cost function.

Question 3.
(i) A manufacturer’s marginal cost function is \(\frac{dC}{dx}\) = \(\frac{500}{\sqrt{2 x+5}}\). If C is in rupees, determine the cost involved to increase production from 100 to 300 units.
(ii) The marginal cost of a manufacturer (in rupees) is given by MC = 0.2x + 3. Determine the cost involved to increase the production from 60 to 70 units.
Solution:
(i) Given Marginal cost function(MC) = \(\frac{dC}{dx}\) = \(\frac{500}{\sqrt{2 x+5}}\)
Required cost involved to increase production from 100 to 300 units = C(300) – C(100)

(ii) Given Marginal cost function (MC) = 0.2x + 3
We know that \(\frac{d \mathrm{C}(x)}{d x}\) = MC
Thus, the required cost involved to increase the production from 60 to 70 units

Question 4.
(i) The marginal cost of a firm is given by MC = 3q² – 4q + 5; q being output, find its output, find its total cost function given that the fixed cost of the firm is ₹ 100.
(ii) The marginal cost of production is found to be MC = 100 – 20x + x², where x is number of units produced. The fixed cost of production of ₹ 90000 . Find the cost function.
(iii) The marginal cost of production is MC = 20 – 0.04x + 0.003x², where x is the number of units produced. The fixed cost of production is ₹ 7000. Find the total cost and average cost function.
Solution:
(i) Given Marginal cost function MC = 3q² – 4q + 5 where q being the output
We know that MC = \(\frac{d}{dq}\)C(q)

given fixed cost of firm be ₹ 100
∴ C(0) = 100
∴ from (1); 100 = K
Thus eqn. (1) reduces to ; C(q) = q³ – 2q² + 5q + 100 be the required total cost function.

(ii) Given Marginal cost function MC = 100 – 20x + x²

It is given that fixed cost of production be ₹ 9000
∴ C(0) = 9000
∴ from (1); 9000 = K
putting the value of K in eqn. (1); we get
C(x) = 100x – 10x² + \(\frac{x^3}{3}\) + 9000 be the required cost function.

(iii) Given marginal cost function (MC) = 20 – 0.04x + 0.003x²
We know that MC = \(\frac{d}{dx}\) C(x)

It is given that the fixed cost of production be ₹ 7000. ….(1)
∴ C(0) = 7000
∴ from (1); 7000 = K
putting the value of K in eqn. (1) ; we have
C(x) = 20x – 0.02x² + 0.001x³ + 7000
be the required cost function.
∴ Average cost function(AC) = \(\frac{C(x)}{x}\) = 20 = 0.02x + 0.001x² + \(\frac{7000}{x}\)

Question 5.
Find the total cost function if
(i) MC = 3(3x + 4)^-1/2 and fixed cost is zero.
(ii) MC = 3000e^0.3x + 50 and fixed cost is ₹ 80,000.
(iii) M C = \(\frac{p}{\sqrt{p x+q}}\), where p, q are constants and fixed cost is zero.
Solution:
(i) Given MC = 5(3x + 4)^-1/2 and MC = \(\frac{dC}{dx}\)

Here fixed cost is 0 ∴ C(0) = 0
∴ from (1); 0 = 2 × 2 + K ⇒ K = -4
Thus eqn. (1) reduces to ; C(x) = \(2 \sqrt{3 x+4}-4\) be the required total cost function.

(ii) Given MC = 3000e^0.3x + 50 and MC = \(\frac{d}{dx}\)C(x)

Since fixed cost is ₹ 80000 ∴ C(0) = 80000
∴ from (1); 80000 = 10000 + K ⇒ K = 70,000
∴ eqn. (1) reduces to ; C(x) = 10000e^0.3x + 50x + 70000 be the required total cost function.

∴ from (1); 0 = 2√q + K ⇒ K = -2√q
putting the value of K in eqn. (1) ; we have
C(x) = 2\(\sqrt{p x+q}\) – 2√q which is the required cost function.

Question 6.
Find the total and average cost functions if MC = 4 – 2x + x², and the fixed cost is ₹ 100.
Solution:
Given MC = 4 – 2x + x² and MC = \(\frac{d}{dx}\)C(x)

Given fixed cost is ₹ 100 ∴ C(0) = 100
∴ from (1); 100 = K
putting the value of K in eqn. (1); we have
C(x) = 4x – x² + \(\frac{x^3}{3}\) + 100 be the required cost function.
∴ Average cost function (AC) = \(\frac{\mathrm{C}(x)}{x}\)
AC(x) = 4 – x + \(\frac{x^2}{3}\) + \(\frac{100}{x}\)

Question 7.
Find the average cost function if
(i) MC = 4 + 3e^x and fixed cost is ₹ 50.
(ii) MC = 2(2x + 25)^-1/2 and fixed cost is ₹ 40.
Solution:
(i) Given MC = 4 + 3e^x and MC = \(\frac{d}{dx}\)C(x)

given fixed cost is ₹ 50 ∴ C(0) = 50
∴ from (1); 50 = 0 + 3 + K ⇒ K = 47
Thus, eqn. (1) reduces to ; C(x) = 4x + 3e^x + 47
∴ Average cost function(AC) = \(\frac{C(x)}{x}\) = 4 + \(\frac{3 e^x}{x}\) + \(\frac{47}{x}\)

Given fixed cost is ₹ 40 ∴ C(0) = 40
∴ from (2); 40 = 2 × 5 + K ⇒ K = 30
Thus eqn. (2) reduces to ; C(x) = 2\(\sqrt{2 x+25}\) + 30 be the required cost function.
∴ Average cost function (AC) = \(\frac{\mathrm{C}(x)}{x}\) = \(\frac{2 \sqrt{2 x+25}}{x}\) = \(\frac{30}{x}\)

Question 8.
(i) The marginal cost function MC for a production is given by MC = \(\frac{2}{\sqrt{4 x+9}}\) and the fixed cost; is ₹ 2000. Find the total cost and the average cost of producing 4 units of the output.
(ii) If the marginal cost function is given by MC = 3(3x + 4)^-1/2 and fixed cost is 2, find the average cost of 7 units of output.
Solution:

It is given fixed cost is ₹ 2000 ∴ C(0) = 2000
∴ from (1); 2000 = 3 + K ⇒ K = 1997
These eqn. (1) reduces to; C(x) = \(\sqrt{4 x+9}\) + 1997
∴ Total cost of producing 4 units of output =C(4) = \(\sqrt{16+9}\) + 1997 = 2002
∴ Average cost function (AC) = \(\frac{\mathrm{C}(x)}{x}\) = \(\frac{\sqrt{4 x+9}}{x}\) + \(\frac{1997}{x}\)
Thus at x = 4 ; \(\frac{\sqrt{16+9}}{4}\) + \(\frac{1997}{4}\) = \(\frac{2002}{4}\) = \(\frac{1001}{2}\) = ₹ 500.50

(ii) Marginal cost function (MC) = 3(3x + 4)^-1/2 and MC = \(\frac{d}{d x}\)C(x)

Given fixed cost is ₹ 2 ∴ C(0) = 2
∴ from (1) ; 2 = 2 × 2 + K ⇒ K = -2
putting the value of K in eqn. (1); we have
C(x) = 2\(\sqrt{3 x+4}\) – 2
∴ Average cost function(AC) = \(\frac{C(x)}{x}\) = \(\frac{2 \sqrt{3 x+4}}{x}\) = \(\frac{2}{x}\)
∴ (AC)_{x = 7} = \(\frac{2 \sqrt{21+4}}{7}\) – \(\frac{2}{7}\) = \(\frac{10}{7}\) – \(\frac{2}{7}\) = \(\frac{8}{7}\)

Question 9.
Given that the marginal cost MC and the average cost AC of a product are directly proportional to each other, find the total cost function so that the cost of producing 2 units is ₹ 8 and of producing 4 units is ₹ 64.
Solution:
Given MC ∝ AC ⇒ MC = KAC where K = constant of proportionality
⇒ \(\frac{dC}{dx}\) = \(\frac{kC}{x}\) ⇒ \(\frac{dC}{C}\) = K\(\frac{dx}{x}\)
On integrating; we have
log C = K log x + log K’ ⇒ log C = log x^K . K’ ⇒ C(x) = x^K K’ ….(1)
It is given that the cost of producing 2 units is ₹ 8.
i.e. when x = 2; C = 8 ∴ from (1); 8 = 2^K . K’ …(2)
also, cost of producing 4 units is ₹ 64.
i.e. x = 4 ; C = 64 ∴ from (1); 64 = 4^K K’ …(3)
on dividing eqn. (3) by eqn. (2); we have

Thus eqn. (1) reduces to ; C(x) = x³ be the required total cost function.