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Worked Examples · Example 5

Q.The total cost C(x)C(x) in Rupees, associated with the production of xx units of an item is given by C(x)=0.005 x3−0.02 x2+30x+5000C(x) = 0.005\,x^3 - 0.02\,x^2 + 30x + 5000. Find the marginal cost when 33 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output.

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Marginal cost is the derivative of the total cost function. Differentiating C(x)C(x) and evaluating at x=3x=3 gives the marginal cost as ₹30.015 per unit.

Why marginal cost?

In economics, "marginal" always means the rate of change — the extra cost of producing one more unit when you're already at a certain output level. For a continuous cost function, that's exactly the derivative dCdx\frac{dC}{dx}. The problem explicitly says "instantaneous rate of change", so we differentiate, not use average cost.

Step-by-step

  1. Write down the cost function

C(x)=0.005x3−0.02x2+30x+5000C(x) = 0.005x^3 - 0.02x^2 + 30x + 5000

  1. Differentiate term by term

    • Derivative of 0.005x30.005x^3: 0.005⋅3x2=0.015x20.005 \cdot 3x^2 = 0.015x^2
    • Derivative of −0.02x2-0.02x^2: −0.02⋅2x=−0.04x-0.02 \cdot 2x = -0.04x
    • Derivative of 30x30x: 3030
    • Derivative of constant 50005000: 00

    So the marginal cost function is

MC(x)=C′(x)=0.015x2−0.04x+30MC(x) = C'(x) = 0.015x^2 - 0.04x + 30

  1. Evaluate at x=3x = 3

MC(3)=0.015(3)2−0.04(3)+30MC(3) = 0.015(3)^2 - 0.04(3) + 30

=0.015⋅9−0.12+30= 0.015 \cdot 9 - 0.12 + 30

=0.135−0.12+30= 0.135 - 0.12 + 30

=0.015+30= 0.015 + 30 …

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