Regular engagement with S Chand Class 12 Maths Solutions Chapter 26 Application of Calculus in Commerce and Economics Ex 26(a) can boost students’ confidence in the subject.
S Chand Class 12 ICSE Maths Solutions Chapter 26 Application of Calculus in Commerce and Economics Ex 26(a)
Question 1.
(i) A manufacturer of steel Almirahs has fixed costs of ₹ 1,00,000 and variable cost of ₹ 400 per unit. Find the equation relating costs to production. What is the cost of producing 100 almirahs?
(ii) A hotel banquets for groups of people at a cost of ₹ 200 per person plus an overhead charge of ₹ 750. Find the cost C(x) they would charge for catering for x people.
(iii) A company finds its cost function to be C(x) = 100 + 50 x and its demand function to be p(x) = 102 – x. Find (a) the revenue function, (b) the profit function.
(iv) A company produced commodity with ₹ 10,000 fixed costs. The variable costs are estimated to 25% of the total revenue received on selling the product at a rate of ₹ 6 per unit. Find the total revenue, total cost and profit functions.
(v) The demand function for a certain commodity is given by p = 1000 – 15x – x2, 0 < x < 25. What is the price per unit and the total revenue from the sale of 2 units?
Solution:
(i) Let x be the number of steel Almirahs
Given TFC = ₹ 100000 and TVC = ₹ 400 × x
∴ TC = TFC + TVC = 100000 + 400x = C(x)
∴ C(100) = 100000 + 400 × 100 = ₹ 140000
(ii) Let x be the number of persons
∴ TVC = 200 × x = ₹ 200x
TFC = overhead charge = ₹ 750
∴ C(x) = TVC + TFC = 200x + 750
(iii) Given C(x) = 100 + 50x and demand function p(x) = 102 – x
∴ Revenue function =R(x) = p(x) × x = (102 – x) x = 102 x – x2
∴ profit function = p(x) = R(x) – C(x) = 102 – x2 – 100 – 50x = 2 – 50x – x2
(iv) Given TFC = ₹ 10,000
Let x be the number of units
R(x) = S.P = ₹ 6x
∴ TVC = 25% of S.P = \(\frac { 1 }{ 4 }\) × 6x = ₹ \(\frac { 3x }{ 2 }\)
Thus C(x) = TFC + TVC = 10000 + \(\frac { 3x }{ 2 }\)
Total Revenue = R(x) = ₹ 6x
∴ profit function = R(x) – C(x) = 6x – 10000 – \(\frac { 3x }{ 2 }\) = ₹ \(\left(\frac{9 x}{2}-10000\right)\)
(v) Given demand function p(x) = 1000 – 15x – x2
∴ Revenue function R(x) = p(x) × x = 1000x – 15x2 – x3
∴R(2) = 1000 × 2 – 15 × 22 – 23 = 2000 – 60 – 8 = ₹ 1932
Thus price of one unit of commodity = ₹ \(\frac { 1932 }{ 2 }\) = ₹ 966
Question 2.
Suppose the cost to produce some commodity is a linear function of output. Find cost as a function of output, if costs are ₹ 4000 for 250 units and ₹ 5000 for 350 units.
Solution:
Given cost function be a linear function of x
Let C(x) =ax + b …(1)
it is given that cost of 250 units be ₹ 4000
∴ from (1); 4000 = 250a + b ….(2)
Also it is given cost of 350 units be ₹ 5000
∴C(x) = 5000 when x = 350
∴ from (1); 5000 = 350a + b
eqn. (3) – eqn. (2) gives ;
1000 = 100a ⇒ a = 10
∴ from (2); b = 4000 – 2500 = 1500
∴ from (1); C(x) =10x + 1500
Question 3.
A manufacturer can sell x items of commodity at price of ₹ (330 – x) each. Find the revenue function. If the cost of producing x items is ₹(x2 + 10x + 12), determine the profit function.
Solution:
Revenue function = R(x) = S.P × x = (330 – x)x = ₹(330x – x2)
given cost function C(x) = ₹ (x2 + 10x + 12)
∴ Profit function = R(x) – C(x) = (330x – x2) – (x2 + 10x + 12) = 320x -2x2 – 12
Question 4.
(i) The fixed cost and the variable cost of x units of a manufactured product of a company are ₹ 4,00,000 and ₹ 80x respectively. If each unit is sold for ₹ 280, what is the breakeven point?
(ii) A textbook publisher finds that the production costs to each book are ₹ 25 and that the fixed costs are ₹ 15,000. If each book can be sold for ₹ 45, determine (a) the cost function, (b) the revenue function, and (c) the break-even point.
(iii) The cost of producing x items per day is given in rupees as C(x) = 2000 + 100√x. If each item can be sold for ₹ 10, determine the break-even point.
(iv) A television manufacturer finds that the total cost for producing and marketing x television sets is C(x)=250 x2 + 325x + 10,000. Each product is sold for ₹ 6,500. Determine the break-even points.
(v) A company has fixed costs of ₹26,000. The cost of producing one item is ₹ 30. If this item sells for ₹ 43, find the break-even point.
(vi) The fixed cost of new product is ₹ 18,000 and the variable cost per unit is ₹ 550. If the demand function is p(x) = 4000 – 150x, find the break-even values.
Solution:
(i) Given TFC = ₹ 4,00,000
TVC = ₹ 80x
∴ TC = C(x) = TFC + TVC = 400000 + 80x
∴R(x) = ₹ 280 × x = ₹ 280x
Thus P(x) = R(x) – C(x) = 280x – 400000 – 80x = 200x – 400000
For break even point ; P(x) = 0 ⇒ 200x – 400000 = 0 ⇒ x = \(\frac{400000}{200}\) =2000
(ii) Given TFC = ₹ 15000
Thus, Cost price of x books = ₹ 25x
∴ TVC = ₹ 25x
Thus C(x) = TFC + TVC = 15000 + 25x
given S.P of one book = ₹ 45
∴ S.P of x books = ₹ 45x
Thus P(x) = 45x
Therefore profit function = P(x) = R(x) – C(x) = 45x – 15000 – 25x = 20x – 15000
For breakeven point ; P(x) = 0 ⇒ 20x – 15000 = 0 ⇒ x = \(\frac{15,000}{20}\) = 750 units
(iii) Given C(x) = 2000 + 100√x
given S.P of each item = ₹ 10
∴ Revenue function R(x) = S.P × x = ₹ 10x
Thus pr function = R(x) -C(x) = 10x – 2000 – 100√x
For break even point ; R(x) = 0 ⇒ 10x -2000 – 100√x = 0
⇒ x – 200 – 10√x = 0
Clearly x = 400 satisfies eqn. (1).
(iv) Given C(x) = 250x2 + 325x + 10000
S.P of each product = ₹ 6500
∴ S.P of x products = R(x) = ₹ 6500x
∴ profit function = P(x) = R(x) – C(x) = 6500x – 250x2 – 325x – 10000 = – 250 x2 + 6175 – 10000
For break even point ; P(x) = 0 ⇒ – 25 (10x2 – 247x + 400) = 0 ⇒ 10x2 – 247x + 400 = 0
x = \(\frac{247 \pm \sqrt{247^2-4 \times 10 \times 400}}{20}\) = \(\frac{247 \pm 212}{20}\) = 23 or 2
(v) Given TFC= ₹ 26000
TVC = ₹ 30x
∴ C(x) = TFC + TVC = 26000 + 30x
S.P of one item = ₹ 43
∴ Revenue function = R(x) = ₹ 43x
∴ profit function = P(x) = R(x) – C(x) = 43x – 26000 – 30x
⇒ P(x) = 13x – 26000
For breakeven point ; P(x) = 0 ⇒ 13x – 26000 = 0 ⇒ x = 2000 units
(vi) Given TFC = ₹ 18,000
TVC = ₹ 550x
∴ C(x) = TFC + TVC = 18000 + 550x
given demand function p(x) = 4000 – 150x
∴ Revenue function = R(x) = p(x) × x = (4000 – 150x)x
Thus profit function = P(x) = R(x) – C(x) = (4000 – 150x)x – 18000 – 550x
= – 150x2 + 3450x – 18000 = – 50(3x2 – 69x + 360)
For breakeven point; P(x) = 0 ⇒ – 50 (3x2 – 69x + 360) = 0 ⇒ 3(x2 – 23x + 120) = 0
⇒ x = \(\frac{23 \pm \sqrt{529-480}}{2}\) = \(\frac{23 \pm 7}{2}\) ⇒ x =15 or 8 units
Question 5.
A profit making company wants to launch a new product. It observes that the fixed cost of the new product is ₹ 35,000 and the variable cost per unit is ₹ 500. The revenue received on the sale of x units is given by 5000 x – 100x. Find (i) profit function (ii) break-even points.
Solution:
(i) Given TFC of product = ₹ 35,000
TVC = ₹(500x), where x = no. of units of product
∴ required cost function = C(x) = 35000 + 500x
given R(x) = 5000x – 100x2
Thus P(x) = R(x) – C(x) = 5000x – 100x2 – 35000 – 500x
⇒ P(x) = 4500x – 100x2 – 35000
(ii) For breakeven point, P(x) = 0 ⇒ – 100(x2 – 45x + 350) = 0
⇒ (x – 35)(x – 10) = 0 ⇒ x = 10,35
(iii) For loss, P(x) < 0 ⇒ x2 – 45x + 350 > 0 ⇒ (x – 10)(x – 35) > 0 ⇒ x > 35 or x < 10
Question 6.
A company has fixed cost of ₹ 10,000 and cost of producing one unit of its product is ₹ 50. If each unit sells for ₹ 75, find the break-even value. Also, find the values of x for which the company results in profit.
Solution:
Given total fixed cost = TFC = ₹ 10000
Ttoal variable cost of producing x units = TVC = ₹ 50x
∴ Total cost function C(x) = TFC + TVC = 10000 + 50x
S.P. of each unit = ₹ 75
∴ R(x) = ₹ 75x
Thus, profit function
P(x) = R(x) – C(x) = 75x – 10000 – 50x ⇒ P(x) = 25x – 10000
For breakeven points P(x) = 0 ⇒ 25x – 10000 = 0 ⇒ x = \(\frac{10000}{25}\) = 400
Thus, the company will remain in profit if it produces and sells more than 400 units of the product.
Question 7.
(i) A company sells its products at ₹ 10 per unit. Fixed costs for the company are ₹ 35,000 and variable costs are estimated to run 30% of total revenue. Determiae the (a) total revenue function
(b) total cost function
(c) quantity the company must sell to cover the fixed cost.
(ii) The price of selling one unit of a product when x units are demanded given by the equation p =4000 – 2x. The fixed costs of the product are ₹ 20,000 and ₹ 1484 per unit are paid for the product to place in a store. Find the level of sales at which the company expects to cover its costs.
Solution:
(i) Let x be the number of products
S.P of one unit = ₹ 10
∴ S.P of x units = R(x) = Revenue function = ₹ 10x
given TFC = 35000
TVC = 30% of R(x) = \(\frac{30}{100}\) × 10x = ₹ 3x
∴C(x) = TFC + TVC = 35000 + 3x
Thus, quantity that the company must sell to cover the fixed cost = \(\frac{35000}{10}\) = 3500
(ii) Given selling price of each unit of product = p = 4000 – 2x
∴ Revenue function = R(x) = px = (4000 – 2x)x
Given TFC = 20000; TVC = 1484x
∴ Total cost function (TC) = C (x) = TFC + TVC ⇒ C(x) = 20000 + 1484x
Thus profit function =P(x) = R(x) – C(x) = (4000 – 2x)x – 20000 – 1484x
= – 2x2 + 2516x – 20000 = -2(x2 – 1258x + 10000)
For breakeven points, P(x) = 0 ⇒ x2 – 1250x + 10000 = 0
⇒ (x – 1250)(x – 8) = 0 ⇒ x = 8, 1250
Hence the company must sell atleast 8 units to cover its costs.
Question 8.
A computer software manufacturer wants to start the production of floppy disks. He observes that he will have to spend ? 2 lakh for the technical know-how. The cost of setting up the machine is ₹ 88,000 and the cost of producing each unit is ? 30. He can sell each floppy at ₹ 45. Determine (i) the total cost function for producing x floppies, and (ii) the break-even point.
Solution:
Given total money spend for technical know how = ₹ 200000
and cost of setting medicine = ₹ 88000
∴ TFC = total fixed cost = 200000 + 88000 = ₹ 288000
TVC = ₹ 30x
∴ C(x) = TFC + TVC = ₹(288000 + 30x)
S.P of one floppy = ₹ 45
∴ S.P of x floppies = R(x) = ₹ 45x
∴ profit function = P(x) = R(x) – C(x) = 45x – 288000 – 30x = 15x – 288000
For breakeven point ; P(x) = 0 ⇒ 15x – 288000 = 0 ⇒ x = \(\frac { 288000 }{ 15 }\) = 19200 units
Question 9.
A company sells its product at the rate of ₹ 6 per unit. The variable costs are estimated to run 25% of the total revenue received. If the fixed costs for the product are ₹ 45,00, find
(i) The total revenue function,
(ii) The total cost function,
(iii) The profit function,
(iv) The break-even point,
(v) The number of units the company must sell to cover its fixed cost.
Solution:
Given Total fixed cost = TFC = ₹ 4500
S.P of one unit = ₹ 6
∴ S.P of x units = ₹ 6x = R(x) = Revenue function
∴ TVC = 25% of R(x) = \(\frac { 1 }{ 4 }\) × 6x = ₹ \(\frac { 3x }{ 2 }\)
Thus C(x) = TFC + TVC = 4500 + \(\frac { 3x }{ 2 }\)
∴ profit function = P(x) = R(x) – C(x) = 6x – 4500 – \(\frac { 3x }{ 2 }\) = ₹\(\left(\frac{9 x}{2}-4500\right)\)
For breakeven point, we put P(x) = 0 ⇒ \(\frac { 9x }{ 2 }\) – 4500 = 0 ⇒ 9x = 9000 ⇒ x = 1000 units
∴ required no. of units the company must sell to cover its fixed cost =\(\frac { 4500 }{ 6 }\) = 750
Question 10.
A company sells x tins of talcum powder each day as ₹ 10 per tin. The cost of mauufacturing is ₹ 6 per tin and the distributor charges ₹ 1 per tin. Besides these daily overhead cost, a fixed cost comes to ₹ 600. Determine the profit function. What is the profit if 500 tins are manufactured and sold a day. How do you interpret the situation if the company manufactures and sells 100 tins a day? What is the break-even point?
Solution:
S.P of one tin of talcum powder = ₹ 10
∴ S.P of x tins of talcum powder = ₹ 10x
∴ Revenue function = R(x) = ₹ 10x
∴ Cost price of one tin of talcum powder = ₹(6 + 1) = ₹ 7
Then total variable cost = TVC = ₹ 7x and TFC = ₹ 600
∴C(x) = TFC + TVC = 600 + 7x
Thus, profit function = P(x) = R(x) – C(x) = 10x – 600 – 7x = 3x – 600
Therefore, profit on manufacturing and selling 500 tins per day = P(500) = 3 × 500 – 600 = ₹ 900
If the company manufactures and sells 100 tins per day.
∴ P(100) = 300 – 600 = -300
Thus the company has ₹ 300 loss.
For break even point ; P(x) = 0 ⇒ 3x – 600 = 0 ⇒ x = 200
Question 11.
From the following information, calculate break-even point in terms of sales value and in units. Fixed factory overhead cost = ₹ 60,000. Fixed selling overhead cost = ₹ 12,000, variable manufacturing cost = ₹ 12 per unit, selling price per unit = ₹ 24.
Solution:
Fixed factory overhead cost = ₹ 60000
fixed selling overhead cost = ₹ 12000
∴ TFC = ₹ 60000 + ₹ 12000 = ₹ 72,000
TVC = ₹ 12x
∴ C(x) = TFC + TVC = 72000 + 12x and R(x) = S.P of x units = ₹ 24x
∴ Profit function = P(x) = R(x) – C(x) = 24x – 72000 – 12x = 12x – 72000
For breakeven point
∴ P(x) = 0 ⇒ 12x – 72000 = 0 ⇒ x = 6000 units
Thus; break even point in terms of sales value = ₹ 6000 × 24 = ₹ 1,44,000
Question 12.
The daily cost of production C for x units of an assembly is given by C(x) = ₹(12.5x + 6400)
(i) If each unit is sold, for ₹ 25, determine the minimum number of units that should be produced and sold to ensure’tio loss.
(ii) If the selling price is reduced by ₹ 2.50 per unit, what would be the break-even point?
(iii) If it is known that 500 units can be sold daily what price per unit should be charged to guarantee no loss?
Solution:
Given cost function C(x)=₹(12.5 x+6400)
(i) given S.P of one unit = ₹ 25
∴ Revenue function = S.P of x units = ₹ 25x
∴ P(x) = R(x) – C(x) = 25x – 12.5x – 6400 = 12.5x – 6400
Thus, the minimum no. of units that should be produced and sold to ensure no loss.
i.e. at breakeven point ∴P(x) = 0 ⇒ 12.5x – 6400 = 0 ⇒ x = \(\frac{6400}{12.5}\) = 512 units
(ii) If S.P of per unit is decreased by ₹ 2.50 per unit
∴ S.P of x units = (25 -2.50)x = ₹ 22.50x
For break even point ; P(x) = 0 ⇒ R(x) – C(x) = 0 ⇒ 22.50x – 12.50x – 6400 = 0
⇒ 10x = 6400 ⇒ x = 640 units
(iii) Since 500 units can be sold daily i.e. x = 500
∴C(500) = (12.5 × 500 + 6400) = ₹ 12,650
Let S.P of each unit be ₹ y
∴ R(x) = S.P of 500 units = ₹ 500y
For no loss i.e. breakeven point; P(x) = 0 ⇒ R(x) = C(x) ⇒ R(500) = C(500)
⇒ 500y = 12,650 ⇒ y = \(\frac{12,650}{500}\) = ₹ 25.30
Hence, required price per unit should be charged to guarantee no loss be ₹ 25.30.
Question 13.
A company decides to set up a small production plant for manufacturing eiectronic clocks. The total cost for initial set up as fixed cost is ₹ 9 lakh. The additional cost (i.e., variable cost) for producing each clock is ₹ 300. Each clock is sold at ₹ 750. During the first month 1500 clocks are produced and sold.
(i) Determine the cost function C(x) for the total cost of producing x clocks.
(ii) Determine the revenue function R(x) for the total revenue from the sale of x clocks.
(iii) Determine the profit function P(x) for the total revenue from the sale of x clocks.
(iv) What profit or loss the company incurs during the first month when all the 1500 clocks are sold.
(v) Determine the break-even point.
Solution:
Given total fixed cost = TFC = ₹ 9lakh
(i) given cost of producing each clock = ₹ 300
∴ Cost of producing x clocks = ₹ 300x = TVC
Thus C(x) = TFC + TVC = 900000 + 300x
(ii) S.P of each clock = ₹ 750
Revenue function = S.P of x clocks =₹ 750 x = R(x)
(iii) ∴ Profit function = P(x) = R(x) – C(x) = 750x – 900000 – 300x = 450x – 900000
(iv) P(1500) = 450 × 1500 – 900000 = 67500 – 900000 = – 2,25,000
Clearly the company incurs a loss of ₹ 2,25,000 during the first month when all the 1500 clocks are sold.
(v) For break-even point ; P(x) = 0 ⇒ 450x – 900000 = 0
∴ x = \(\frac{900000}{450}\) = 2000 units
Question 14.
The pricing policy of a company follows the demand function p = D(x), D(x) being the price per unit when x units are demanded. After studying the market trends the company determines the price function that is given by D(x) = 2000 – 4x.
If the product is to be marketed the company will incur a fixed cost of ₹ 60,000 and will have to pay ₹ 600 for each unit that is produced and placed in the store. At what sales level can the company expect to recover its costs?
Solution:
Given demand function D(x) = 2000 – 4x
∴Revenue function = R(x) = D(x) × x = (2000 – 4x)x = 2000x – 4x2
given TFC = ₹ 60000
TVC = ₹ 600x
C(x) = TFC + TVC = 60000 + 600x
Since at break even point ; R(x) = C(x)
⇒ 2000x – 4x2 = 60000 + 600x ⇒ 4 x2 – 1400x + 60000 = 0
⇒ x2 – 350x + 15000 = 0 & ⇒ (x – 300)(x – 50) = 0 ⇒ x = 300,50