Effective S Chand Class 12 Maths Solutions Chapter 26 Application of Calculus in Commerce and Economics Ex 26(d) can help bridge the gap between theory and application.
S Chand Class 12 ICSE Maths Solutions Chapter 26 Application of Calculus in Commerce and Economics Ex 26(d)
Question 1.
The demand function for a manu- facturers product is p = 20 – \(\frac { x }{ 4 }\), where x is the number of units and p is the price per unit. At what value of x will there be revenue? What is the maximum revenue?
Solution:
Given demand function is given by p = 20 – \(\frac { x }{ 4 }\)
∴ Revenue function = R(x) = px = \(\left(20-\frac{x}{4}\right) x\) x = 20x – \(\frac{x^2}{4}\)
Now \(\frac { d }{ x }\)R(x) = 20 – \(\frac { 2x }{ 4 }\) and \(\frac{d^2}{d x^2}\)R(x) = –\(\frac { 2 }{ 4 }\) = – \(\frac { 1 }{ 2 }\)
For maxima/minima, \(\frac { d }{ dx }\)R(x) = 0 ⇒ 20 – \(\frac { x }{ 2 }\) = 0 ⇒ x = 40
at x = 40; \(\frac{d^2}{d x^2}\)R(x) = – \(\frac { 1 }{ 2 }\) < 0
Hence Revenue is maximise when x = 40 units
∴ Maximum revenue function = 20 × 40 – \(\frac{(40)^2}{4}\) = 800 – 400 = 400
Question 2.
The unit dem and function is x = \(\frac { 1 }{ 4 }\) (24 – 2p), where x is the number of units demanded and p is the price per unit. Find (i) the revenue function R in terms of price p. (ii) the price and number of units demanded for which the revenue is maximum.
Solution:
Given demand function is x = \(\frac { 1 }{ 3 }\)(24 – 2p) ⇒ 3x = 24 – 2p ⇒ p = \(\frac{24-3 x}{2}\) …(1)
Thus, revenue function R(x) = px = \(\left(12-\frac{3}{2} x\right) x\)
∴ \(\frac { d }{ dx }\)R(x) = 12 – 3x and \(\frac{d^2}{d x^2}\)R(x) = – 3
For maxima/minima, we put \(\frac { d }{ dx }\)R(x) = 0 ⇒ 12 – 3x = 0 ⇒ x = 4
∴ \(\left[\frac{d^2}{d x^2} \mathrm{R}(x)\right]_{x=4}\) = -3 < 0
Thus Revenue function R is maximise when x = 4 units
∴ from (1); p = \(\frac { 24 – 12 }{ 2 }\) = ₹ 6 and Revenue function = px = \(\frac { 1 }{ 3 }\) (24p – 2p2)
Question 3.
The demand function for a particular commodity is y = 15 e-x / 3 for 0 ≤ x ≤ 8, where p is the price per unit and x is the number of units demanded. Determine the price and quantity for which total revenue is maximum.
Solution:
Given demand function is y = 15 e-x / 3 for 0 ≤ x ≤ 8
∴ revenue function R(x) = yx = 15 e-x / 3 . x
Question 4.
A club has 1000 menbers who are each paying ₹ 100 per month. The club proposes to increase the monthly membership fee and it is expected that for every increase of ₹ 1, five members will discontinue the service. Find what increase will yield maximum revenue and what will this revenue be?
Solution:
Let ₹ x be the required increase in monthly fee.
Since it is given that for each increase of ₹ 1, 5 persons will discontinue the membership.
Thus new no. of members = (1000 – 5x) and new rate of membership = (100 + x)
∴ revenue function = R(x) = (100 + x)(1000 – 5x)
∴ \(\frac { dR }{ dx }\) = (100 + x)(- 5) + (1000 – 5x) = 500 – 10x and \(\frac{d^2 \mathrm{R}}{d x^2}\) = – 10
For maxima/minima, we put \(\frac { dR }{ dx }\) = 0 ⇒ 500 – 10x = 0 ⇒ x = 50
∴ \(\left(\frac{d^2 \mathrm{R}}{d x^2}\right)_{x=50}\) = 10 < 0
Thus, R is maximum when x = 50
∴ Required increase in monthly for be ₹ 50.
and Max. Revenue = (100 + 50)(1000 – 250) = 150 × 750 = 1,12,500
Question 5.
A company charges ? 550 for a transister set on orders of 50 or less sets. The charge is reduced by ₹ 5 per set for each set ordered in excess of 50. Find the largest size order company should allow so as to receive maximum revenue.
Solution:
Let x be the required excess of transistor sets.
Since it is given that the charge is reduced by ₹ 5 per set for each set ordered in excess of 50 .
Then new no. of sets = (50 + x)
new price of transistor = (550 – 5x)
∴ revenue function R(x) = (50 + x)(550 – 5x)
∴ \(\frac { dR }{ dx }\) = (50 + x) (-5) + (550 – 5x) . 1 = 300 – 10x and \(\frac{d^2 \mathrm{R}}{d x^2}\) = – 10
For maxima/minima, we put \(\frac { dR }{ dx }\) = 0 ⇒ 300 – 10x = 0 ⇒ x = 30
at x = 30; \(\frac{d^2 \mathrm{R}}{d x^2}\) = 10 < 0
Thus R is maximum when x =30
∴ required largest size order that the company should allow so as to receive maximum revenue will be (50 + 30) i.e. 80 sets.
Question 6.
A steel plant is capable of producing x tonnes per day of a low grade steel and y tonnes per day of a nigh grade steel, where y = \(\frac{42-5 x}{10-x}\). If the fixed market price of low grade steel is half of high grade steel, show that 6 tonnes of low grade steel are produced per day for maximum total revenue.
Solution:
Given y = \(\frac{42-5 x}{10-x}\) where x tonnes of low grade steel and y tonnes of high grade steel.
∴ revenue function R(x) = yx = \(\left(\frac{42-5 x}{10-x}\right)\)x
For maxima/minima, \(\frac { dR }{ dx }\) = 0 ⇒ x2 – 20x + 84 = 0 ⇒ (x – 6) (x – 14) = 0 ⇒ x = 16, 14
at x = 6; \(\frac{d^2 \mathrm{R}}{d x^2}\) = \(\frac{160}{(6-10)^3}\) = –\(\frac{160}{64}\) = –\(\frac{5}{2}\) < 0
Thus R is maximum when x = 6 Hence 6 tonnes of low grade steel are produced per day for maximum total revenue.
Question 7.
A tour operator engages a bus of seating capacity of 50 for taking children on an excursion and charges ₹ 100 per child with an additional charge of ₹ 2.50 for each subsequent cancellation. Determine the total revenue R(x), as a function of x, the number of cancellation received prior to the departure date. What is the value of x for which R(x) maximum.
Solution:
Let x be the number of cancellations received prior to the departure date.
Since it is given that a tour operator engages a bus of seating capacity of 50 children and charges ₹ 100 per child with an additional charge of ₹ 2.50 for each subsequent cancellation.
∴ new no. of children =50 – x and new charge = ₹ (100 + 2.5x)
Thus, total revenue function R(x) = (50 – x)(100 + 2.5x)
⇒ R(x) = 5000 + 25x – \(\frac { 5 }{ 2 }\)x2 ∴ \(\frac { dR }{ dx }\) = 25 – 5x and \(\frac{d^2 \mathrm{R}}{d x^2}\) = -5
For maxima/minima, \(\frac { dR }{ dx }\) = 0 ⇒ 25 – 5x = 0 ⇒ x = 5 and \(\left(\frac{d^2 \mathrm{R}}{d x^2}\right)_{x=5}\) = -5 < 0
Thus R is maximise when x = 5
∴ required no. of cancellations = 5
Question 8.
The revenue function of a product is given by the relation y = 4000000 – (x – 2000)2, where y is the total revenue and x is the number of units sold. Find (i) the number of units sold which maximizes total revenue, (ii) the amount of maximum revenue.
Solution:
Given total revenue function y = 4000000 – (x – 2000)2
∴ \(\frac { dy }{ dx }\) = -2(x – 2000) and \(\frac{d^2 y}{d x^2}\) = -2
For maxima/minima, \(\frac { dy }{ dx }\) = 0 ⇒ x – 2000 = 0 ⇒ x = 2000
at x = 2000; \(\frac{d^2 y}{d x^2}\) = -2 < 0
Thus y is maximise when x = 2000
Thus total revenue is maximise when x = 2000
and Maximum revenue = R(2000) = 4000000 – (2000-2000)2 = ₹ 4000000
Question 9.
The total cost and demand functions of an item are given by C(x) = \(\frac{x^3}{3}\) – 7x2 + 111x + 50 and p = 100 – x respectively. Write the total revenue function and the profit function. Find the profit maximizing level of output x and the maximum profit.
Solution:
Given cost function C(x) = \(\frac{x^3}{3}\) – 7x2 + 111x + 50 and demand function p = 100 – x
∴ Total revenue function = R(x) = px = (100 – x)x
and profit function = P(x) = R(x) – C(x)
∴ P(x) = 100x – x2 – \(\frac{x^3}{3}\) + 7x2 – 111x – 50 = –\(\frac{x^3}{3}\) + 6x2 – 11x = 50
\(\frac{d}{dx}\)P(x) = -x2 + 12x – 11 and \(\frac{d^2}{d x^2}\)P(x) = -2x + 12
For maxima/minima, \(\frac{dP}{dx}\) = 0 ⇒ -x2 + 12x – 11 = 0 ⇒ x2 – 12x + 11 = 0
⇒ (x – 1)(x – 11) = 0 ⇒ x = 1, 11
and \(\left(\frac{d^2}{d x^2} \mathrm{P}(x)\right)_{x=11}\) = – 22 + 12 = – 10 < 0
Thus profit function P(x) is maximum when x = 11
and Maximum profit = P (11) = –\(\frac{11^3}{3}\) + 6 × 112 – 121 – 50
= –\(\frac{1331}{3}\) + 726 – 171 = \(\frac{-1331+2178-513}{3}\) = \(\frac{343}{3}\)
Question 10.
A company has produced x items and the total cost C and total revenue R are given by the equation C = 100 + 0.05 x2 and R = 3x. Find how many items should be produced to maximize the profit. What is this profit?
Solution:
Given cost function C(x) = 100 + 0.015x2 and Revenue function R = 3x
Let P(x) = profit function = R(x) – C(x) = 3x – 100 – 0.015x2
∴\(\frac{d}{dx}\)P(x) = 3 – 0.03x and \(\frac{d^2}{d x^2}\)P(x) = -0.03
For maxima/minima, \(\frac{d}{dx}\)P(x) = 0 ⇒ 3 – 0.03x = 0 ⇒ x = \(\frac{3}{0.03}\) = 100
∴ at x = 100; \(\frac{d^2}{d x^2}\)P(x) = -0.03 < 0
Thus, profit P is maximum when x = 100
and hence required no. of items to get maximum profit be 10 .
Also maximum profit =P(100) = 300 – 100 – 0.015 × 10000 = 300 – 100 – 150 = 50
Question 11.
A radio manufacturer finds that he can sell x radios per week at ₹ p each, where p = 2\(\left(100-\frac{x}{4}\right)\). His cost of production of x radios per week is ₹ \(\left(120 x+\frac{x^2}{2}\right)\). Show that his profit is maximum when the production is 40 radios per week. Find also the maximum profit per week.
Solution:
Given p = 2\(\left(100-\frac{x}{4}\right)\)
∴ Revenue function R(x) = px = 2\(\left(100-\frac{x}{4}\right)\)x
and cost function C(x) = ₹\(\left(120 x+\frac{x^2}{2}\right)\)
Let P(x) be the profit function
Then P(x) = R(x) – C(x) = 200x – \(\frac{x^2}{2}\) – 120x – \(\frac{x^2}{2}\) = 80x – x2
∴ \(\frac{d}{dx}\)P(x) = 80 – 2x and \(\frac{d^2}{d x^2}\)P(x) = -2
For maxima/minima, \(\frac{d}{dx}\)P(x) = 0 ⇒ 80 – 2x = 0 ⇒ x = 40
at x = 40; \(\frac{d^2}{d x^2}\)P(x) = – 2 < 0
Thus P(x) is maximise when x = 40
Hence profit is maximum when the production is 40 radios per week.
∴ Max. profit per week = P(40) = 80 × 40 – 402 = 3200 – 1600 = ₹ 1600
Question 12.
(i) Find the profit maximizing output level given x = 200 – 10p and AC = 10 + \(\frac{x}{25}\), where x represents the units of output, p represents price, and AC represents average cost.
(ii) The demand function of an output is x = 106 – 2p, where x is the number of units of output and; the price per unit output. If the total revenue be px, determine the number of units for maximum profit.
Solution:
(i) Given x = 200 – 10p ⇒10p = 200 – x ⇒ p = 20 – \(\frac{x}{10}\)
∴ Revenue function R(x) = px = \(\left(20-\frac{x}{10}\right) x\)
given AC = 10 + \(\frac{x}{25}\), Also AC = \(\frac{\mathrm{C}(x)}{x}\)
∴ C(x) = AC × x = \(\left(10+\frac{x}{25}\right) x\)
Thus profit function P(x) = R(x) – C(x) = \(\left(20-\frac{x}{10}\right) x\) – \(\left(10+\frac{x}{25}\right) x\)
= 10x – \(\frac{x^2}{10}\) – \(\frac{x^2}{25}\) = 10x – \(\frac{7 x^2}{50}\)
∴ \(\frac{d}{dx}\)P(x) = 10 – \(\frac{14 x}{50}\) and \(\frac{d^2}{d x^2}\)P(x) = – \(\frac{14}{50}\)
For maxima/minima, we put \(\frac{d}{dx}\)P(x) = 0 ⇒ 10 – \(\frac{14x}{50}\) = 0 ⇒ x = \(\frac{250}{7}\)
and \(\frac{d^2}{d x^2}\)P(x) = – \(\frac{7}{25}\) < 0 at x = \(\frac{250}{7}\)
Thus profit P(x) is maximise when x = \(\frac{250}{7}\)
(ii) Given demand function is given by x = 106 – 2p ⇒ p = \(\frac{106-x}{2}\) = 53 – \(\frac{x}{2}\)
∴ Revenue function R(x) = px = \(\left(53-\frac{x}{2}\right) x\) and C(x) = 7x
∴ profit function P(x) = R(x) – C(x) = 53x – \(\frac{x^2}{2}\) – 7x = 46x – \(\frac{x^2}{2}\)
∴ \(\frac{d}{dx}\)P(x) = 46 – x and \(\frac{d^2}{d x^2}\)P(x) = -1
For maxima/minima, we put \(\frac{d}{dx}\)P(x) = 0 ⇒ 46 – x = 0 ⇒ x = 46
and \(\frac{d^2}{d x^2}\)P(x) = – 1 < 0
Thus P(x) is maximise when x = 46 units
and maximum profit = 46 × 46 – \(\frac{(46)^2}{2}\) = 46 × 23 = ₹ 1058
Question 13.
The cost function C(x) for producing x units of a commodity is given by
C(x) = \(\frac{1}{3}\) x3 – 5x2 + 75x + 10 .
At what level of output the marginal cost \(\left(\text { i.e., } \frac{d C}{d x}\right)\) attains its mininum? What is the marginal cost at this level of production?
Solution:
Given cost function C(x) = \(\frac{x^3}{3}\) – 5x2 + 75x + 10
∴ Marginal cost function MC = \(\frac{dC}{dx}\) = x2 – 10x + 75
Now \(\frac{d}{dx}\)(MC) = 2x – 10 and \(\frac{d^2}{d x^2}\)(MC) = 2
For maxima/minima, \(\frac{d}{dx}\)(MC) = 0 ⇒ 2x – 10 = 0 ⇒ x = 5
at x = 5; \(\frac{d^2}{d x^2}\)(MC) = 2 > 0
Thus marginal cost function is minimise when x = 5
∴ Minimum marginal cost = 52 – 50 + 75 = 50
Question 14.
The total cost function of producing and marketing x units of a commodity is given by C = 16 – 12x + 2x2. Find the level of output at which it is minimum.
Solution:
Given cost function C = 16 – 12x + 2x2
∴ \(\frac{d \mathrm{C}}{d x}\) = – 12 + 4x and \(\frac{d^2 \mathrm{C}}{d x^2}\) = 4
For maxima/minima, \(\frac{d \mathrm{C}}{d x}\) = 0 ⇒ – 12 + 4x = 0 ⇒ x = 3
and \(\left(\frac{d^2 \mathrm{C}}{d x^2}\right)_{x=3}\) = 4 > 0
Thus C(x) is minimise when x = 3
Hence the required level of output be 3 units.
Question 15.
The total cost function of a product is given by C(x) = x3 – 315x2 + 27,000x + 20,000 where x is the number of units produced. Determine the number of units that should be produced to minimize the total cost.
Solution:
Given total cost function C(x) = x3 – 315x2 + 27000x + 20000
∴ \(\frac{d}{dx}\)C(x) = 3x2 – 630x + 27000 and \(\frac{d^2}{d x^2}\)C(x) = 6x – 630
For maxima/minima, \(\frac{d}{dx}\)C(x) = 0 ⇒ 3(x2 – 210x + 9000) = 0
⇒ (x – 150)(x – 60) = 0 ⇒ x= 150, 60
at x = 150; \(\frac{d^2}{d x^2}\)C(x) = 900 – 630 = 270 > 0
Thus cost is minimise when x = 150 units.
Hence the required no. of units that should be produced to minimize the total cost be 150 units.
Question 16.
The manufacturing cost of an item consists of ₹ 1000 as overheads, material cost ₹ 2 per unit x2 and the labour cost ₹\(\frac{x^2}{90}\) for x units produced. Find how many units of the item should be produced so that the average cost is minimum.
Solution:
Total fixed cost = TFC = ₹ 1000
given material cost per unit = ₹ 2
∴ material cost for x units = ₹ 2x
Total labour cost =₹ \(\left(2 x+\frac{x^2}{70}\right)\)
Thus, total cost function C(x) = TFC + TVC = 1000 + 2x + \(\frac{x^2}{90}\)
∴ Average cost (AC) = \(\frac{\mathrm{C}(x)}{x}\) = \(\frac{1000}{x}\) + 2 + \(\frac{x}{90}\)
∴ \(\frac{d}{dx}\)(AC) = – \(\frac{1000}{x^2}\) + \(\frac{1}{90}\) and \(\frac{d^2}{d x^2}\)(AC) = \(\frac{2000}{x^3}\)
For maxima/minima, \(\frac{d}{dx}\)(AC) = 0
⇒ \(\frac{-1000}{x^2}+\frac{1}{90}=0\) ⇒ x2 = 9000 = (300)2 ⇒ x = 300 (∵ x > 0)
at x = 300; \(\frac{d}{dx}\)(AC) = \(\frac{2000}{(300)^3}\) > 0
Thus AC i.e. average cost is minimum when x = 300 units
Question 17.
The cost function of a firm is C = 5x2 + 28x + 5, where C is the cost and $x$ is level of output. A tax at the rate of ₹ 2 per unit of output is imposed and the producer adds it to his cost. Find the minimum value of the average cost.
Solution:
Given cost function C = 5x2 + 28x + 5
∴ Total tax imposed on x unit of output = ₹ 2x
∴ Total cost function = C(x) = 5x2 + 30x + 5
∴ Average cost function (AC) = \(\frac{\mathrm{C}(x)}{x}\) = 5x + 30 + \(\frac{5}{x}\)
∴ \(\frac{d}{dx}\)(AC) = 5 – \(\frac{5}{x^2}\) and \(\frac{d^2}{d x^2}\)(AC) = \(\frac{10}{x^3}\)
For maxima/minima, \(\frac{d}{dx}\)(AC) = 0 ⇒ 5 – \(\frac{5}{x^2}\) = 0 ⇒ x = 1 (∵ x > 0)
∴ \(\frac{d^2}{d x^2}\)(AC) = \(\frac{10}{1^3}\) = 10 > 0
Thus average cost is minimise when x = 1 unit and min average cost = 5 + 30 + 5 = 40
Question 18.
The manufacturing cost of an article involves a fixed overhead of ₹ 100 per day ₹ 0.50 for x2 material and \(\frac{x^2}{100}\) per day for labour and machinery to produce x articles. How many articles should be produced per day to minimize the average cost per article.
Solution:
Given TFC = fixed overhead cost = ₹ 100
TVC = 0.50 × x + \(\frac{x^2}{100}\)
Thus average cost (AC) = \(\frac{\mathrm{TC}}{x}\) = \(\frac{100}{x}\) + \(\frac{1}{2}\) + \(\frac{x}{100}\)
∴ \(\frac{d}{dx}\)(AC) = – \(\frac{100}{x^2}\) + \(\frac{1}{100}\)
∴ \(\frac{d^2}{d x^2}\)(AC) = \(\frac{200}{x^3}\)
For maxima/minima, \(\frac{d}{dx}\) AC = 0 ⇒ \(\frac{-100}{x^2}\) + \(\frac{1}{00}\) = 0 ⇒ x = 100
and \(\frac{d^2}{d x^2}\)(AC) = \(\frac{200}{(100)^3}\) = \(\frac{2}{10000}\) > 0
Thus AC is minimise when x = 100 units
Question 19.
(i) A firm produces x units of output per week at a total cost of ₹ \(\frac{x^3}{3}\) – x2 + 5x + 3, find the output levels at which the marginal cost and average variable cost attain their respective minima.
(ii) The cost function of a firm is given by C = \(\frac{1}{3}\)x3 – 5x + 30x + 10, where C is the total cost for x items. Determine x at which the marginal cost is minimum.
Solution:
(i) Given total cost function C(x) = \(\frac{x^3}{3}\) – x2 + 5x + 3
∴ Marginal cost function (MC) = \(\frac{d \mathrm{C}}{d x}\) = x2 – 2x + 5
and \(\frac{d}{dx}\)(MC) = 2x – 2 and \(\frac{d^2}{d x^2}\)(MC) = 2
For maxima/minima, \(\frac{d}{dx}\)(MC) = 0 ⇒ 2x – 2 = 0 ⇒ x = 1
and \(\left[\frac{d^2}{d x^2}(\mathrm{MC})\right]_{x=1}\) = 2 > 0
Thus marginal cost (MC) is minimise when x = 1
Hence the required output level at which MC is minimum be 1 unit.
From eqn. (1); we have
Total variable cost = \(\frac{x^3}{3}\) – x2 + 5x and TFC = c(0) = 3
Thus, AVC = \(\frac{\text { TVC }}{x}\) = \(\frac{x^2}{3}\) – x + 5
∴ \(\frac{d}{dx}\)(AVC) = \(\frac{2 x}{3}\) – 1 and \(\frac{d^2}{d x^2}\)(AVC) = \(\frac{2}{3}\)
For maxima/minima, \(\frac{d}{dx}\)(AVC) = 0 ⇒ \(\frac{2x}{3}\) = 1 ⇒ x = \(\frac{3}{2}\)
and x = \(\frac{3}{2}\); \(\frac{d^2}{d x^2}\)(AVC) = \(\frac{2}{3}\) > 0
Thus AVC is minimise when x = \(\frac{3}{2}\) units
(ii) Given C(x) = \(\frac{x^3}{3}\) – 5x2 + 30x + 10
∴ Marginal cost function (MC) = \(\frac{dC}{dx}\) = x2 – 10x + 30
∴ \(\frac{d}{dx}\)(MC) = 2x – 10 and \(\frac{d^2}{d x^2}\)(MC) = 2
For maxima/minima, we put \(\frac{d}{dx}\)(MC) = 0 ⇒ 2x – 10 = 0 ⇒ x = 5
∴ at x = 5; \(\frac{d^2}{d x^2}\)(MC) = 2 > 0
This MC is minimise when x = 5 units
Question 20.
Given a quadratic cost function C(x) = ax2 + bx + c, minimize the average cost and show that the average cost is equal to marginal cost at that value.
Solution:
Given cost function C(x) = ax2 + bx + c
∴ average cost (AC) = \(\frac{C(x)}{x}\) = ax + b + \(\frac{c}{x}\)
Now \(\frac{d}{dx}\)(AC) = a – \(\frac{c}{x^2}\) and \(\frac{d^2}{d x^2}\)(AC) = \(\frac{2 c}{x^3}\)
Hence the average cost is equal to marginal cost at the point of minimum average cost.