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Application of Calculus - Cost, Revenue, and Profit Functions

Grade 12ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Total Cost Function (C(x)C(x)) consists of a Fixed Cost (FCFC), which is constant regardless of production level, and a Variable Cost (VC(x)VC(x)), which changes with the number of units (xx) produced. The graph typically starts at the value of FCFC on the y-axis.

Graph showing Total Cost curve starting from Fixed Cost intercept.
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The Average Cost (ACAC) is the cost per unit, calculated as C(x)x\frac{C(x)}{x}. Geometrically, the value of xx that minimizes ACAC occurs at the point where the Marginal Cost (MCMC) curve intersects the ACAC curve from below.

Graph showing the intersection of Marginal Cost and Average Cost at the minimum point of AC.
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Marginal Revenue (MRMR) is the rate of change of Total Revenue (R(x)R(x)) with respect to the quantity sold (xx). It represents the additional revenue generated by selling one more unit.

Parabolic Total Revenue curve showing revenue increasing then decreasing.
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Profit Maximization occurs when the difference between R(x)R(x) and C(x)C(x) is at its greatest positive value. This happens when MR=MCMR = MC and the second derivative of the profit function is negative (P′′(x)<0P''(x) < 0).

Flowchart for finding the profit-maximizing level of output.
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Break-even point is the level of production where Total Revenue equals Total Cost (R(x)=C(x)R(x) = C(x)), resulting in zero profit. At this point, the business covers all its expenses but does not yet earn a surplus.

Intersection of Revenue and Cost lines representing the break-even point.

📐Formulae

Total Cost: C(x)=FC+VC(x)C(x) = FC + VC(x)

Average Cost: AC=C(x)xAC = \frac{C(x)}{x}

Marginal Cost: MC=ddx[C(x)]MC = \frac{d}{dx}[C(x)]

Total Revenue: R(x)=p⋅xR(x) = p \cdot x (where pp is the demand function)

Marginal Revenue: MR=ddx[R(x)]MR = \frac{d}{dx}[R(x)]

Profit Function: P(x)=R(x)−C(x)P(x) = R(x) - C(x)

Average Profit: AP=P(x)xAP = \frac{P(x)}{x}

Marginal Profit: MP=dPdx=MR−MCMP = \frac{dP}{dx} = MR - MC

Condition for Profit Maximization: MR=MCMR = MC and d2Pdx2<0\frac{d^2P}{dx^2} < 0

💡Examples

Problem 1:

The cost function for a manufacturer is given by C(x)=13x3−5x2+30x+10C(x) = \frac{1}{3}x^3 - 5x^2 + 30x + 10. Find the Marginal Cost and Average Cost when x=10x = 10.

Solution:

Step 1: Find Marginal Cost (MCMC). MC=dCdx=ddx(13x3−5x2+30x+10)MC = \frac{dC}{dx} = \frac{d}{dx}(\frac{1}{3}x^3 - 5x^2 + 30x + 10) MC=x2−10x+30MC = x^2 - 10x + 30 At x=10x = 10, MC=(10)2−10(10)+30=100−100+30=30MC = (10)^2 - 10(10) + 30 = 100 - 100 + 30 = 30.

Step 2: Find Average Cost (ACAC). AC=C(x)x=13x3−5x2+30x+10x=13x2−5x+30+10xAC = \frac{C(x)}{x} = \frac{\frac{1}{3}x^3 - 5x^2 + 30x + 10}{x} = \frac{1}{3}x^2 - 5x + 30 + \frac{10}{x} At x=10x = 10, AC=13(10)2−5(10)+30+1010AC = \frac{1}{3}(10)^2 - 5(10) + 30 + \frac{10}{10} AC=1003−50+30+1=33.33−19=14.33AC = \frac{100}{3} - 50 + 30 + 1 = 33.33 - 19 = 14.33.

Explanation:

We use the derivative of the cost function to find the marginal cost and the ratio of total cost to units to find the average cost.

Problem 2:

A company sells xx items at a price of p=200−2xp = 200 - 2x each. The cost of producing xx items is C(x)=40x+1200C(x) = 40x + 1200. Determine the value of xx that maximizes the profit.

Solution:

Step 1: Find the Revenue function R(x)R(x). R(x)=p⋅x=(200−2x)x=200x−2x2R(x) = p \cdot x = (200 - 2x)x = 200x - 2x^2

Step 2: Find the Profit function P(x)P(x). P(x)=R(x)−C(x)=(200x−2x2)−(40x+1200)P(x) = R(x) - C(x) = (200x - 2x^2) - (40x + 1200) P(x)=−2x2+160x−1200P(x) = -2x^2 + 160x - 1200

Step 3: Find the derivative dPdx\frac{dP}{dx} and set it to zero for critical points. P′(x)=ddx(−2x2+160x−1200)=−4x+160P'(x) = \frac{d}{dx}(-2x^2 + 160x - 1200) = -4x + 160 Set P′(x)=0  ⟹  −4x+160=0  ⟹  4x=160  ⟹  x=40P'(x) = 0 \implies -4x + 160 = 0 \implies 4x = 160 \implies x = 40

Step 4: Check the second derivative for maximization. P′′(x)=ddx(−4x+160)=−4P''(x) = \frac{d}{dx}(-4x + 160) = -4 Since P′′(x)<0P''(x) < 0, the profit is maximized at x=40x = 40.

Explanation:

To maximize profit, we first construct the profit function from revenue and cost, then find the level of output where the first derivative is zero and ensure the second derivative is negative.

Problem 3:

A manufacturer's demand function is p=500−5xp = 500 - 5x and the cost function is C(x)=100x+2000C(x) = 100x + 2000. Calculate the Marginal Revenue (MRMR) and Marginal Cost (MCMC) when x=20x = 20 units. Also, determine if the profit is increasing or decreasing at this level of output.

Graph of MR and MC lines where MR is above MC at x=20.

Solution:

  1. Find Revenue: R(x)=p×x=(500−5x)x=500x−5x2R(x) = p \times x = (500 - 5x)x = 500x - 5x^2
  2. Find MRMR: MR=dRdx=500−10xMR = \frac{dR}{dx} = 500 - 10x. At x=20x = 20, MR=500−10(20)=300MR = 500 - 10(20) = 300
  3. Find MCMC: MC=dCdx=100MC = \frac{dC}{dx} = 100
  4. Find Marginal Profit: MP=MR−MC=300−100=200MP = MR - MC = 300 - 100 = 200 Since MP>0MP > 0 at x=20x = 20, the profit is increasing.

Explanation:

We differentiate the Revenue and Cost functions to find the marginal values. Because the Marginal Revenue is higher than the Marginal Cost, each additional unit produced adds more to income than to expense, thus increasing total profit.

Problem 4:

The total cost of producing xx units is C(x)=5x2+20x+500C(x) = 5x^2 + 20x + 500. Find the level of output xx at which the Average Cost (ACAC) is minimized.

U-shaped Average Cost curve showing the minimum point at x=10.

Solution:

  1. Find Average Cost: AC=C(x)x=5x2+20x+500x=5x+20+500xAC = \frac{C(x)}{x} = \frac{5x^2 + 20x + 500}{x} = 5x + 20 + \frac{500}{x}
  2. Differentiate ACAC with respect to xx: d(AC)dx=5−500x2\frac{d(AC)}{dx} = 5 - \frac{500}{x^2}
  3. Set derivative to zero for minimum: 5−500x2=0  ⟹  5x2=500  ⟹  x2=100  ⟹  x=105 - \frac{500}{x^2} = 0 \implies 5x^2 = 500 \implies x^2 = 100 \implies x = 10
  4. Verify using second derivative: d2(AC)dx2=1000x3\frac{d^2(AC)}{dx^2} = \frac{1000}{x^3}, which is positive for x=10x=10, so ACAC is minimized.

Explanation:

To find the minimum average cost, we first derive the ACAC function by dividing the total cost by the number of units. We then find the stationary point by setting the derivative to zero. The second derivative test confirms it is a minimum.