ML Aggarwal Class 12 Maths Solutions Section C Chapter 04 Applications Of Derivatives In Commerce And Economics

Applications of Derivatives in Commerce and Economics

Introduction

Mathematical and quantitative methods are increasingly used to tackle business and economic challenges. Differential calculus helps find how quickly a function (the dependent variable) changes when one of the independent variables shifts. Integration, being the reverse of differentiation, involves determining a function when its rate of change is already known.

This chapter begins with key economic ideas - fixed costs, variable costs, average cost, revenue, and profit - and then moves into marginal functions (marginal cost and marginal revenue) using the first derivative. Second derivatives will help us locate minimum costs and find where revenue or profit reaches its highest point.

4.1 Fixed and Variable Cost; Average Cost

Production cost depends on several factors - plant size, output level, raw material prices, and technology.

\( C = f(S, O, P, T, \ldots) \)

In this chapter, we look at cost as depending only on output level. So the total cost to produce x units is:

\( TC = f(x) \)

Total cost breaks down into two parts - fixed costs and variable costs.

Fixed Costs are amounts paid regardless of how much is made - examples include interest, rent, and permanent employee wages. So:

\( TFC = TC \text{ when } x = 0 \)

Fixed cost stays the same whether production goes up or down.

Variable costs move with output - raw materials and casual labour wages are common examples. So:

\( \text{Total Cost} = \text{Total Fixed Cost} + \text{Total Variable Cost} \)

\( TC = TFC + TVC \)

The TFC is steady while TVC begins at zero when output is zero and increases from there. TC is the sum of TFC and TVC. In practice, TVC typically does not follow a straight line. Initially it climbs at a slowing rate, reaches an inflection point, then climbs at an accelerating rate. This becomes clearer when we examine marginal ideas.

Average cost comes from dividing total cost by the output level.

\( C = f(x) \implies AC = \frac{C}{x} = \frac{f(x)}{x} \)

Also:

\( TC = TFC + TVC \)

\( \implies \frac{TC}{x} = \frac{TFC + TVC}{x} \)

\( \implies AC = AFC + AVC \)

So Average Cost = Average Fixed Cost + Average Variable Cost

When variable cost stays constant, the average cost curve looks like figure (i) below. But in real situations, economies of scale only hold up to a point, after which the cost of each new unit starts climbing again. Then the average cost curve appears as shown in figure (ii) below.

Illustrative Examples

Example 1. For manufacturing a certain item, the fixed cost is Rs. 6000 and the cost of producing each unit is Rs. 20.
(i) What is the cost function?
(ii) What is the total cost and average cost of producing 15 units?
(iii) What is the total cost and average cost of producing 100 units?
Solution: (i) Total fixed cost = Rs. 6000,
Variable cost of producing one unit = Rs. 20
\( \therefore \) Total cost, TC = Rs. (6000 + 20x), where x is the number of units produced.

(ii) Total cost of producing 15 units = \( TC|_{x=15} \)
\( = \) Rs. (6000 + 20 \times 15) = Rs. 6300
\( \therefore \) Average cost of producing 15 units = Rs. \( \frac{6300}{15} \) = Rs. 420.

(iii) Total cost of producing 100 units = Rs. (6000 + 20 \times 100) = Rs. 8000
\( \therefore \) Average cost of producing 100 units = Rs. \( \frac{8000}{100} \) = Rs. 80.

Example 2. The cost function for a certain commodity is \( C(x) = 3 + 2x - \frac{1}{4}x^2 \). Write the various cost components (TC, TFC, TVC, AC, AFC, AVC) when 4 items are produced. Verify your result.
Solution: Total cost TC or \( C(x) = 3 + 2x - \frac{1}{4}x^2 \)

Total fixed cost, \( TFC = C(x)|_{x=0} = 3 \)

Total variable cost, \( TVC = 2x - \frac{1}{4}x^2 \)

Average cost, \( AC = \frac{C(x)}{x} = \frac{3}{x} + 2 - \frac{1}{4}x \)

Average fixed cost, \( AFC = \frac{TFC}{x} = \frac{3}{x} \)

Average variable cost, \( AVC = \frac{TVC}{x} = 2 - \frac{1}{4}x \)

When x = 4, we get:

Total Cost = \( C(4) = 3 + 2 \times 4 - \frac{1}{4} \times 4^2 = 7 \)

Total fixed cost TFC = 3

Total variable cost TVC = \( 2 \times 4 - \frac{1}{4} \times 4^2 = 4 \)

We see that TC = TFC + TVC

Average Cost \( AC = \frac{3}{4} + 2 - \frac{1}{4} \times 4 = 1\frac{3}{4} \)

Average fixed cost \( AFC = \frac{3}{4} \)

Average variable cost \( AVC = 2 - \frac{1}{4} \times 4 = 1 \)

We see that AC = AFC + AVC.

Example 3. It is known that cost of producing 100 units of a commodity is Rs. 250 and cost of producing 200 units is Rs. 300. Assuming that AVC is constant, find the cost function.
Solution: Let the total fixed cost TFC be a.
Let AVC = constant = b
Then TVC = (AVC) x = bx
\( \therefore \) TC = TFC + TVC = a + bx

Given that when x = 100, TC = Rs. 250 and when x = 200, TC = Rs. 300
\( \therefore \) a + 100 b = 250 \( \ldots (i) \)
a + 200 b = 300 \( \ldots (ii) \)

Subtracting (i) from (ii),
100 b = 50 \( \implies b = \frac{1}{2} \)

Putting this value of b in (i),
\( a + 100 \times \frac{1}{2} = 250 \implies a = 200 \)

Hence, the cost function is \( C(x) = 200 + \frac{1}{2}x \).

Exercise 4.1

Question 1. For manufacturing a certain item, the fixed cost is Rs. 6500 and the cost of producing each unit is Rs. 12.50.
(i) What is the cost function? Draw a graph, clearly indicating TC, TFC and TVC.
(ii) What is the total cost of producing 75 items?
(iii) What is the average cost of producing 400 items?
Answer: (i) \( C(x) = 6500 + 12.50x \)

(ii) Rs. 7437.50

(iii) Rs. 28.75
In simple words: Multiply the fixed cost and the per-unit cost by the number produced, then add them. To find average cost, divide the total cost by how many units you made.

Exam Tip: Make certain you can separate fixed costs (which never change) from variable costs (which grow with production) and know how to sketch all three lines on your graph.

Question 2. The cost function for a certain commodity is \( C(x) = 12 + 3x - \frac{1}{3}x^2 \). Write down the total cost, fixed cost, variable cost and average cost when 3 units are produced.
Answer: Total cost TC = \( C(3) = 12 + 3(3) - \frac{1}{3}(3)^2 = 12 + 9 - 3 = 18 \), so TC = Rs. 18

Fixed cost TFC = \( C(0) = 12 \), so TFC = Rs. 12

Variable cost TVC = TC - TFC = 18 - 12 = Rs. 6

Average cost AC = \( \frac{TC}{x} = \frac{18}{3} = 6 \), so AC = Rs. 6
In simple words: Work out the total cost using the formula, then take away the fixed part to get the variable part. Average cost is just the total cost split equally across all units.

Exam Tip: Remember that TFC is found by setting x = 0 in the cost function, and TVC is everything remaining after you remove the fixed part.

Question 3. Cost of producing 75 units of a commodity is Rs. 275 and cost of producing 150 units is Rs. 300. Assuming that TVC is linear, find the:
(i) cost function.
(ii) average total cost of producing 75, 150, 225 units respectively.
(iii) average fixed cost of producing 75, 150, 225 units respectively.
(iv) average variable cost of producing 75, 150, 225 units respectively.
Answer: (i) Since TVC is linear, we have TC = TFC + TVC = a + bx for constants a and b.

From the given data:
a + 75b = 275 - (i)
a + 150b = 300 - (ii)

Subtracting (i) from (ii): 75b = 25, so b = 1/3
From (i): a = 275 - 75(1/3) = 275 - 25 = 250

So the cost function is \( C(x) = 250 + \frac{1}{3}x \)

(ii) Average total cost = TC/x
At x = 75: AC = 275/75 = Rs. 3.67
At x = 150: AC = 300/150 = Rs. 2.00
At x = 225: AC = \( 250 + \frac{1}{3}(225) \) / 225 = 325/225 = Rs. 1.44

(iii) Average fixed cost = TFC/x = 250/x
At x = 75: AFC = 250/75 = Rs. 3.33
At x = 150: AFC = 250/150 = Rs. 1.67
At x = 225: AFC = 250/225 = Rs. 1.11

(iv) Average variable cost = TVC/x = (1/3)x/x = 1/3
At x = 75: AVC = Rs. 0.33
At x = 150: AVC = Rs. 0.33
At x = 225: AVC = Rs. 0.33
In simple words: Use two known points to find the two unknowns in your linear formula. Average fixed cost drops as you make more, but average variable cost stays the same because it is constant.

Exam Tip: Verify your cost function by checking that both given data points satisfy the equation you found.

Question 4. The total cost of producing and marketing x units of commodity is given by \( C = 2x + e^x + 5e \). Find (i) the fixed cost (ii) the variable cost (iii) total cost of producing 5 units (iv) average cost of producing 5 units.
Answer: (i) Fixed cost = \( C(0) = 2(0) + e^0 + 5e = 1 + 5e = \) Rs. \( 5e \)

(ii) Variable cost = TC - FC = \( 2x + e^x + 5e - 5e = 2x + e^x \) = Rs. \( (2x + e^x) \)

(iii) Total cost at x = 5: TC\( |_{x=5} = 2(5) + e^5 + 5e = 10 + e^5 + 5e \)

(iv) Average cost at x = 5: AC = \( \frac{10 + e^5 + 5e}{5} = 2 + \frac{e^5}{5} + e \)
In simple words: The fixed cost is what you pay when you make zero units. The variable cost grows with each unit you produce. Divide total cost by units to get the average.

Exam Tip: Always set x = 0 to find fixed cost - this separates out the part that does not depend on production level.

4.2 Demand Function; Revenue and Profit Functions

4.2.1 Demand Function

Research in economics shows that how much of a product people want depends on many things - the product's price, what buyers earn, their likes, and prices of similar items. To keep things simple, we only look at the link between how much people want and what the price is, treating everything else as unchanging.

Usually price per unit and quantity demanded move in opposite directions - when price goes up, quantity demanded falls, or when more is wanted, the price comes down.

Demand function \( x = f(p) \)

By custom, price p is plotted on the vertical axis against quantity demanded x on the horizontal axis.

Thus, \( p = f(x) \)

Remarks

• If how much people want depends on the item's price, the demand function is written as \( x = f(p) \). For example, \( x = 70 + 3p \) means p is the price per unit and x is how many units are demanded.
• If the price relies on how much demand there is, the demand function is written as \( p = f(x) \). For example, \( p = 30 - 5x + 2x^2 \) means x units are demanded and p is the price per unit.
• From the above, any connection between x and p counts as a demand function. We can rearrange to make x the subject or make p the subject.

4.2.2 Revenue Function

Revenue means the money a company gets by selling a fixed amount of a product. Let p equal the price per unit and x equal how many units are sold. Then total revenue is:

\( R \text{ or } R(x) = p \cdot x \)

When price p is fixed, then R(x) forms a straight line naturally. When price p moves with demand x, then \( p = f(x) \), so:

\( R \text{ or } R(x) = (\text{demand}) \cdot (\text{price}) = xf(x) \)

The average revenue, or revenue per unit, is given by:

\( AR = \frac{R}{x} = \frac{px}{x} = p \)

So average revenue equals the price per unit.

4.2.3 Profit Function

The profit function P(x) or \( \pi(x) \) for making, selling, and marketing x units of a product is:

\( P(x) \text{ or } \pi(x) = R(x) - C(x) \)

where R(x) is the revenue function and C(x) is the cost function.

The average profit is:

\( \frac{P(x)}{x} = \frac{R(x)}{x} - \frac{C(x)}{x} \)

So average profit = average revenue - average cost.

When the government imposes taxes, profit goes down; when it gives subsidies, costs drop and profit goes up.

4.3 Breakeven Analysis

Usually, when businesses spend on fixed capital costs, they lose money when they make or sell little. But as output and sales climb, the average cost falls, and after a point, the business starts earning profit. This brings us to breakeven point study.

The breakeven point is the production level where sales revenue matches total production cost. At this stage, the business earns no profit and takes no loss.

So at breakeven point,

\( R(x) = C(x) \)

i.e. \( R(x) - C(x) = 0 \)

i.e. \( P(x) = 0 \)

Any of these forms can be used to find the breakeven point. There may be zero, one, or many breakeven points.

Illustrative Examples

Example 1. (i) A chocolate bar sells for Rs. 20. What is the total revenue and average revenue by selling 30 bars?
(ii) The demand function for T.V. sets is p = 20000 - 100x (rupees). Determine the total revenue and average revenue by selling 20 sets.
Solution: (i) Price p = Rs. 20
Total revenue by selling 30 bars = Rs. (30 × 20) = Rs. 600
Average revenue by selling 30 bars = R/x = Rs. 600/30 = Rs. 20

(ii) Price p = Rs. (20000 - 100x) per item
\( \therefore \) Total revenue by selling x sets = px = Rs. (20000 - 100x)x
\( \therefore \) Total revenue by selling 20 sets = Rs. (20000 - 100 × 20) × 20 = Rs. (18000 × 20) = Rs. 360000
Average revenue by selling 20 items = Rs. 360000/20 = Rs. 18000
In simple words: Multiply the price by how many you sell to get total revenue. Divide the total by the quantity to get revenue per item.

Exam Tip: Keep track of whether price is fixed or varies with demand - this changes how you compute revenue.

Example 2. The price of a commodity is fixed at Rs. 55 and its cost function is C(x) = 30x + 250.
(i) Determine the breakeven point.
(ii) What is the profit when 12 items are sold?
(iii) What is the profit when 5 items are sold?
Solution: (i) Here revenue R(x) = (price)(demand) = 55x,
Cost C(x) = 30x + 250
\( \therefore \) Profit function P(x) = R(x) - C(x) = 55x - (30x + 250) = 25x - 250

To determine the breakeven point, we have P(x) = 0
\( \implies \) 25x - 250 = 0 \( \implies \) x = 10

Hence, the breakeven point is x = 10. At this level of production, revenue = cost = Rs. 550 and profit = 0.

(ii) When 12 items are produced,
Profit = Rs. (25 × 12 - 250) = Rs. (300 - 250) = Rs. 50
Thus, there is a profit of Rs. 50 when 12 items are produced and sold.

(iii) When 5 items are sold,
Profit = Rs. (25 × 5 - 250) = Rs. (125 - 250) = - Rs. 125
The negative sign shows that there is a loss.
Thus, there is a loss of Rs. 125 when only 5 items are produced and sold.
In simple words: When profit equals zero, you have hit the breakeven point - you make back what you spent but earn nothing extra. More units bring profit, fewer units bring loss.

Exam Tip: Always solve P(x) = 0 to find the breakeven point, and check whether production below or above this level gives profit or loss.

Example 3. A manufacturer finds that selling price of a product is Rs. 25 while cost is C(x) = 30x + 120. What would you advise him?
Solution: Here revenue R(x) = (price)(demand) = 25x
Cost C(x) = 30x + 120
\( \therefore \) Profit P(x) = R(x) - C(x) = 25x - (30x + 120) = -5x - 120

We can see that profit is always negative, regardless of the level of production.

Alternatively, to determine breakeven point, we have P(x) = 0
\( \implies \) -5x - 120 = 0 \( \implies \) x = -24

As the number of units \( \geq \) 0, we see that there is no breakeven point.

As the project is doomed from the start, the businessmen should not take up this project.
In simple words: The cost per unit is too high compared to the selling price. No matter how many you make, you will always lose money.

Exam Tip: If the breakeven point comes out negative, the project cannot work - abandon it.

Example 4. The fixed cost of a new product is Rs. 18000 and the variable cost per unit is Rs. 550. If the demand function is p(x) = 4000 - 150x, find the breakeven values.
Solution: Let x units of the product be produced and sold.
As the variable cost per units is Rs. 550,
\( \therefore \) the variable cost of producing x units = Rs. 550x.
As the fixed cost is Rs. 18000,
\( \therefore \) total cost of producing x units, C(x) = Rs. (18000 + 550x).

Given demand function is p(x) = 4000 - 150x i.e. the selling price per unit is Rs. (4000 - 150x).
\( \therefore \) Total revenue on selling x units,
R(x) = (price per unit)(number of units sold) = Rs. (4000 - 150x)x

At breakeven values, C(x) = R(x)
\( \implies \) 18000 + 550x = (4000 - 150x)x
\( \implies \) 150x² - 4000x + 550x + 18000 = 0
\( \implies \) 150x² - 3450x + 18000 = 0
\( \implies \) x² - 23x + 120 = 0
\( \implies \) (x - 8)(x - 15) = 0
\( \implies \) x = 8, 15

Hence, the breakeven values are x = 8 and x = 15.
In simple words: Set revenue equal to cost and solve. Two breakeven points mean the company breaks even at two different production levels - profit happens in between.

Exam Tip: When demand varies with price, expect a quadratic equation and likely two breakeven points.

Example 5. The total cost and the total revenue of a company that produces and sells x units of particular product are respectively C(x) = 5x + 350 and R(x) = 50x - x².
Find (i) the breakeven values (ii) the values of x that produce a profit (iii) the values of x that result in a loss.
Solution: (i) At breakeven values, R(x) = C(x)
\( \implies \) 50x - x² = 5x + 350 \( \implies \) x² - 45x + 350 = 0
\( \implies \) (x - 10)(x - 35) = 0 \( \implies \) x = 10 or 35
Hence, the breakeven values are x = 10 and x = 35.

(ii) For profit, R(x) > C(x) \( \implies \) 50x - x² > 5x + 350
\( \implies \) x² - 45x + 350 < 0 \( \implies \) (x - 10)(x - 35) < 0
\( \implies \) 10 < x < 35

(iii) From (i) and (ii) we see that losses occur when x < 10 or x > 35.

Alternatively, losses occur when R(x) < C(x)
\( \implies \) 50x - x² < 5x + 350 \( \implies \) x² - 45x + 350 > 0
\( \implies \) (x - 10)(x - 35) > 0 \( \implies \) x < 10 or x > 35
In simple words: Between the two breakeven points, you make profit. Below the first or above the second, you lose money.

Exam Tip: Always factor the inequality to find the ranges where profit > 0 and where profit < 0.

Example 6. A company produces a commodity with Rs. 24000 fixed cost. The variable cost is estimated to be 25% of the total revenue recovered on selling the product at a rate of Rs. 8 per unit. Find the following:
(i) Cost function (ii) Revenue function (iii) Break-even point.
Solution: Let x units of the product be produced and sold. As the selling price of one unit is Rs. 8, so the total revenue on selling x units = Rs. 8x.

(i) Since the variable cost is 25% of total revenue recovered,
so the variable cost = 25% of Rs. 8x = Rs. \( \left(\frac{25}{100} \times 8\right)x \) = Rs. 2x.

Fixed cost of the company is Rs. 24000.
\( \therefore \) Cost function (in Rs.) = C(x) = 2x + 24000.

(ii) Revenue function (in Rs.) = R(x) = 8x.

(iii) At break-even points, R(x) = C(x)
\( \implies \) 8x = 2x + 24000 \( \implies \) 6x = 24000 \( \implies \) x = 4000.
Hence, break-even point is x = 4000.
In simple words: Work out 25% of the selling revenue to get the variable cost. Set revenue equal to the sum of all costs to find breakeven.

Exam Tip: When variable cost is a percentage of revenue, express it as a fraction and simplify before setting up the breakeven equation.

Example 7. A company is selling a certain product. The demand function for the product is linear. The company can sell 2000 units when the price is Rs. 8 per unit and it can sell 3000 units when the price is Rs. 4 per unit. Determine:
(i) the demand function (ii) the total revenue function.
Solution: We are given two points on the demand curve: (2000, 8) and (3000, 4).

Since the demand function is linear, we have p = a + bx, where a and b are constants.

From the given points:
8 = a + 2000b ... (i)
4 = a + 3000b ... (ii)

Subtracting (ii) from (i):
4 = -1000b \( \implies \) b = -0.004

From (i): 8 = a + 2000(-0.004) = a - 8 \( \implies \) a = 16

(i) Therefore, the demand function is p = 16 - 0.004x

(ii) Total revenue function R(x) = (price)(quantity) = (16 - 0.004x)x = 16x - 0.004x²
In simple words: Use the two given points to set up a system of equations and solve for the constants in the linear demand formula.

Exam Tip: When setting up a linear function with two points, use the slope formula: slope = (y₂ - y₁)/(x₂ - x₁).

Answers to Exercises

Exercise 4.1

1. (i) C(x) = 6500 + 12.50x; (ii) Rs. 7437.50; (iii) Rs. 28.75

2. TC = Rs. 18, AC = Rs. 6

3. (i) C(x) = 250 + (1/3)x; (ii) Rs. 3.67, Rs. 2, Rs. 1.44; (iii) Rs. 3.33, Rs. 1.67, Rs. 1.11; (iv) Rs. 0.33, Rs. 0.33, Rs. 0.33

4. (i) FC = 5e; (ii) VC = 2x + eˣ; (iii) TC|ₓ₌₅ = 10 + e⁵ + 5e; (iv) AC|ₓ₌₅ = 2 + (e⁵/5) + e

Exercise 4.2

1. (i) P(x) = 10x - 150; breakeven point is x = 15; (ii) P(x) = -5x - 150; no breakeven point; (iii) P(x) = 56x - 4x² - 180; breakeven at x = 5, 9; (iv) P(x) = 650x - 5x² - 20000; breakeven at x = 50, 80

2. (i) C(x) = 15000 + 30x; (ii) R(x) = 45x; (iii) x = 1000

3. x = 5, 9

4. (i) C(x) = 150000 + 150x; (ii) R(x) = 350x; (iii) P(x) = 200x - 150000; (iv) x = 750

5. P(x) = 67x - 40200, breakeven point is 600

6. (i) R(x) = 6x; (ii) C(x) = 4500 + 1.5x; (iii) P(x) = 4.5x - 4500; (iv) x = 1000; (v) 750 units

7. (i) 579; (ii) 440; (iii) Rs. 3.15

8. (i) P(x) = 4500x - 100x² - 35000; (ii) x = 10, 35; (iii) x < 10 or x > 35

9. Breakeven points are x = 40, 320. Thus, the company should produce minimum 40 units to recover its cost.

10. 400 units; the company will always remain in profit if it produces and sells more than 400 units of the product.

11. Breakeven points are x = 8, 1250. Hence, the company must sell at least 8 units to cover its costs.

12. (i) R(x) = 5x; (ii) C = 3200 + 1.25x; (iii) x = 853.33; (iv) 640 units

13. P(x) = 5x - 1200; Rs. 3800 profit; Rs. 1300 profit; Rs. 200 loss

14. (i) C = (20/9)(245 - 7p); (ii) R = (p/9)(245 - 7p); (iii) P = (1/9)(p - 20)(245 - 7p); (iv) p = 20, x = 105/9

Exercise 4.3

1. (i) AC = (1200/x) + 20 + x; (ii) MC = 20 + 2x; (iii) MC = 40 at x = 10; it indicates that Rs. 40 are needed to increase the production from 10 to 11 units; (iv) Rs. 41; it indicates exact amount needed to increase the production from 10 to 11 units

2. (i) MC = x² + 6x - 16; (ii) AC = (1/3)x² + 3x - 16 + (2/x)

3. (i) 16.5; (ii) 36

4. (i) x² + 6x - 7; (ii) (x²/3) + 3x - 7 + (16/x)

5. (i) x² + 6x - 7

8. (i) Total cost = x² + 5x + 6, marginal cost = 2x + 5; (ii) x > 6

9. (i) MC = 3x² - 48x + 600, AC = x² - 24x + 600