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

complete question

The demand and total cost functions of a good are

4P+Q-16=0

TC=4+2Q-3Q²/10+Q³/20

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

2:Solve the equation   dr/dQ=0 and hence determine the value of Q which maximizes profit.

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

solution

(1)$TR=P\times Q$

$=[4-\frac{Q}{4}]\times Q$

$=4Q-\frac{Q^{2}}{4}$

$MR=\frac{\delta TR}{\delta Q}$

$MR=\frac{\delta 4Q-\frac{Q^{2}}{4}}{\delta Q}$

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

$(profit)\pi=TR-TC$

$=[4Q-\frac{Q^{2}}{4}]-[4+2Q-\frac{3Q^{2}}{10}+\frac{Q^{3}}{20}]$

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

$=2Q-\frac{Q^{2}}{4}+\frac{5Q^{2}+6Q^{2}}{20}-\frac{Q^{3}}{20}-4$

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

$MC=\frac{\partial TC}{\partial Q}$

$=\partial\div \partial Q[4+2Q-\frac{3Q^{2}}{10}+\frac{Q^{3}}{20}]$

$MC=2-\frac{6Q}{10}+\frac{3Q^{2}}{20}$

(2) $\frac{\partial \pi}{\partial Q}=\pi^{"}=0$

$=\partial\div \partial Q[2Q+\frac{Q^{2}}{20}-\frac{Q^{3}}{20}-4]=0$

$2+\frac{2Q}{20}-\frac{3Q^{2}}{20}=0$

$40+2Q-3Q^{2}=0$

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

$3Q^{2}-12Q+10Q-40=0$

$3Q(Q-4)+10(Q-4)=0$

$(3Q+10)(Q-4)=0$

$Q=\frac {-10}{3}$

$Q=4$

Since Q could never be negative therefore,-10/3 is rejected.

Hence=4 maximizes profit.

(3)

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

at Q=4 $MR=4-\frac{1}{2}(4)=2$

$MC=2-\frac{6Q}{10}+\frac{3Q^{2}}{20}$

at Q=4 $MC=2-\frac{6\times4}{10}+\frac{3\times 4^{2}}{20}=2$

at Q=4

$MR=MC=2$