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Worked Examples · Example 4

Q.The total cost (in ₹) of producing xx units of a product is C(x)=x3−6x2+15x+50C(x) = x^3 - 6x^2 + 15x + 50. Find the average cost and the marginal cost when x=5x = 5, and interpret the marginal cost.

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Average cost.

C‾(x)=C(x)x=x3−6x2+15x+50x=x2−6x+15+50x.\overline{C}(x) = \frac{C(x)}{x} = \frac{x^3 - 6x^2 + 15x + 50}{x} = x^2 - 6x + 15 + \frac{50}{x}.

At x=5x = 5:

C‾(5)=25−30+15+505=25−30+15+10=20.\overline{C}(5) = 25 - 30 + 15 + \frac{50}{5} = 25 - 30 + 15 + 10 = 20.

So the average cost is ₹20 per unit.

Marginal cost.

MC=dCdx=3x2−12x+15.\text{MC} = \frac{dC}{dx} = 3x^2 - 12x + 15.

At x=5x = 5:

MC(5)=3(25)−12(5)+15=75−60+15=30.\text{MC}(5) = 3(25) - 12(5) + 15 = 75 - 60 + 15 = 30.

So the marginal cost is ₹30.

Interpretation. The marginal cost of ₹30 at x=5x = 5 estimates the additional cost of producing the 6th unit. As a check of meaning, the actual extra cost of the 6th unit is C(6)−C(5)C(6) - C(5):

C(6)=216−216+90+50=140,C(5)=125−150+75+50=100,C(6) = 216 - 216 + 90 + 50 = 140, \qquad C(5) = 125 - 150 + 75 + 50 = 100, …

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