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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

Total Cost Function C(x)C(x): This represents the total expense incurred by a company to produce xx units of a commodity. It consists of Fixed Cost (FCFC), which remains constant regardless of production level, and Variable Cost (VCVC), which changes with the number of units. Visually, the Total Cost curve typically starts at a point on the y-axis representing the fixed cost and rises as xx increases.

Average Cost (ACAC): This is the cost per unit of production, calculated as AC=C(x)xAC = \frac{C(x)}{x}. Visually, the Average Cost curve is generally U-shaped; it initially declines as fixed costs are spread over more units (economies of scale) and eventually rises due to inefficiencies at very high production levels.

Marginal Cost (MCMC): This represents the instantaneous rate of change of the total cost with respect to the number of units produced. Mathematically, MC=dCdxMC = \frac{dC}{dx}. In a graph, the Marginal Cost at any point xx is the slope of the tangent to the Total Cost curve at that point.

Revenue Function R(x)R(x): This is the total amount received from the sale of xx units of a product. If pp is the price per unit (demand function), then R(x)=pxR(x) = p \cdot x. Visually, the Revenue curve often starts at the origin (0,0)(0,0) because zero sales result in zero revenue.

Marginal Revenue (MRMR): This is the rate of change of total revenue with respect to the number of units sold, given by MR=dRdxMR = \frac{dR}{dx}. It represents the approximate revenue generated by selling one additional unit. Graphically, it is the slope of the Total Revenue curve.

Profit Function P(x)P(x): Profit is the difference between total revenue and total cost, expressed as P(x)=R(x)C(x)P(x) = R(x) - C(x). The Break-even point occurs when R(x)=C(x)R(x) = C(x) or P(x)=0P(x) = 0. Visually, these are the points where the Revenue and Cost curves intersect.

Profit Maximization: A firm achieves maximum profit at a production level where Marginal Revenue equals Marginal Cost (MR=MCMR = MC) and the second derivative of the profit function is negative (d2Pdx2<0\frac{d^2P}{dx^2} < 0). Visually, this corresponds to the highest peak on the Profit curve graph.

📐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)=pxR(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=MRMCMP = \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)=13x35x2+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(13x35x2+30x+10)MC = \frac{dC}{dx} = \frac{d}{dx}(\frac{1}{3}x^3 - 5x^2 + 30x + 10) MC=x210x+30MC = x^2 - 10x + 30 At x=10x = 10, MC=(10)210(10)+30=100100+30=30MC = (10)^2 - 10(10) + 30 = 100 - 100 + 30 = 30.

Step 2: Find Average Cost (ACAC). AC=C(x)x=13x35x2+30x+10x=13x25x+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)25(10)+30+1010AC = \frac{1}{3}(10)^2 - 5(10) + 30 + \frac{10}{10} AC=100350+30+1=33.3319=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=2002xp = 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)=px=(2002x)x=200x2x2R(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)=(200x2x2)(40x+1200)P(x) = R(x) - C(x) = (200x - 2x^2) - (40x + 1200) P(x)=2x2+160x1200P(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+160x1200)=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.