Question

The demand function for a monopolist's product is
*p=1300-7q* and the average cost per unit for producing q
units is
*c**=0.004**q**2**-1.6q+100+**5000**/q*

-Find the quantity that minimizes the average cost function and the corresponding price. Interpret your results.

-What are the quantity and the price that maximize the profit? What is the maximum profit? Interpret your result.

Answer #1

c = 0.004q^^{2}- 1.6q + 100 + (5,000 / q)

p = 1,300 - 7q

**a)** Average cost = (c / q) = 0.004q - 1.6 + (100
/ q) + (5,000 / q^^{2})

Average cost is minimum when first derivative of average cost with respect to q is 0.

First derivative of Average cost = 0.004 - (100 /
q^^{2}) + (-10,000 / q^^{3})

0.004 - (100 / q^^{2}) + (-10,000 / q^^{3}) =
0

q = 195

At q = 195, p = 1,300 - 7 * 195 = -65

**b)** Profit = Total Revenue - Total Cost

Total Revenue = p * q = 1,300q - 7q^^{2}

Total Cost = 0.004q^^{2}- 1.6q + 100 + (5,000 / q)

Profit = 1,300q - 7q^^{2} - 0.004q^^{2} + 1.6q -
100 - (5,000 / q)

First derivative of profit = 1,300 - 14q - 0.008q + 1.6 - (5,000
/ q^^{2})

1,300 - 14q - 0.008q + 1.6 - (5,000 / q^^{2}) = 0

q = 9

Profit = 1,113.8 at this q level.

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