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Worked Examples · Example 7

Q.A firm faces the demand p=50−xp = 50 - x (price in ₹ when xx units are sold) and total cost C(x)=100+20xC(x) = 100 + 20x. Find the output that maximises profit, the maximum profit, and the price charged. Verify that MR = MC at this output.

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Revenue. R(x)=p x=(50−x)x=50x−x2R(x) = p\,x = (50 - x)x = 50x - x^2.

Profit.

π(x)=R(x)−C(x)=(50x−x2)−(100+20x)=30x−x2−100.\pi(x) = R(x) - C(x) = (50x - x^2) - (100 + 20x) = 30x - x^2 - 100.

Maximise. Differentiate and set to zero:

π′(x)=30−2x=0 ⇒ x=15.\pi'(x) = 30 - 2x = 0 \ \Rightarrow\ x = 15.

Second derivative: π′′(x)=−2<0\pi''(x) = -2 < 0, so x=15x = 15 gives a maximum.

Maximum profit.

π(15)=30(15)−(15)2−100=450−225−100=125.\pi(15) = 30(15) - (15)^2 - 100 = 450 - 225 - 100 = 125.

So the maximum profit is ₹125.

Price charged. From the demand, p=50−x=50−15=35p = 50 - x = 50 - 15 = 35, i.e. ₹35 per unit.

Verify MR = MC (dual check).

MR=dRdx=50−2x=50−2(15)=20,MC=dCdx=20.\text{MR} = \frac{dR}{dx} = 50 - 2x = 50 - 2(15) = 20, \qquad \text{MC} = \frac{dC}{dx} = 20. …

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