Answer to Question #195640 in Calculus for Emmanuella

Question

The demand and total cost functions of a good are respectively and

Find expressions for TR, (profit) , MR, and MC in terms of Q.

Solve the equation

and hence determine the value of Q which maximizes profit.

Verify that, at the point of maximum profit, MR=MC.

Answer 1

The demand and total cost functions of a good are

$4P+Q-16=0$

and

$TC=4+2Q-\dfrac{3Q^2}{10}+\dfrac{Q^3}{20}$

respectively.

a)

$4P+Q-16=0$

$P=4-\dfrac{1}{4}Q$

$TR=P\cdot Q=(4-\dfrac{1}{4}Q)\cdot Q=4Q-\dfrac{Q^2}{4}$

$TR=4Q-\dfrac{Q^2}{4}$

$T\pi=TR-TC$

$T\pi=4Q-\dfrac{Q^2}{4}-(4+2Q-\dfrac{3Q^2}{10}+\dfrac{Q^3}{20})$

$T\pi=-4+2Q+\dfrac{Q^2}{20}-\dfrac{Q^3}{20}$

$MR=\dfrac{d(TR)}{dQ}=\dfrac{d}{dQ}(4Q-\dfrac{Q^2}{4})=4-\dfrac{Q}{2}$

$MR=4-\dfrac{Q}{2}$

$MC=\dfrac{d(TC)}{dQ}=\dfrac{d}{dQ}(4+2Q-\dfrac{3Q^2}{10}+\dfrac{Q^3}{20})$

$=2-\dfrac{3Q}{5}+\dfrac{3Q^2}{20}$

b) Solve the equation

$\dfrac{d(T\pi)}{dQ}=0$

$\dfrac{d(T\pi)}{dQ}=\dfrac{d}{dQ}(-4+2Q+\dfrac{Q^2}{20}-\dfrac{Q^3}{20})$

$=2+\dfrac{Q}{10}-\dfrac{3Q^2}{20}$

$2+\dfrac{Q}{10}-\dfrac{3Q^2}{20}=0, Q\geq 0$

$3Q^2-2Q-40=0$

$D=(-2)^2-4(3)(40)=484=22^2$

$Q=\dfrac{2\pm\sqrt{22^2}}{2(3)}=\dfrac{1\pm11}{3}$

$Q_1=-\dfrac{10}{3}, Q_2=4$

Since $Q\geq 0$

If $0\leq Q<4, \dfrac{d(T\pi)}{dQ}>0, T\pi$ increases.

If $Q>4, \dfrac{d(T\pi)}{dQ}<0, T\pi$ decreases.

The function $T\pi$ has a local maximum at $Q=4.$

Since the function $T\pi$ has the only extremum for $Q\geq 0,$ then the function $T\pi$ has the absolute maximum at $Q=4$ for $Q\geq 0.$

c) $Q=4$

$MR(4)=4-\dfrac{4}{2}=2$

$MC(4)=2-\dfrac{3(4)}{5}+\dfrac{3(4)^2}{20}=2$

Then at the point of the maximum profit $(Q=4)$

$MR(4)=2=MC(2)$