Answer to Question #304822 in Financial Math for James

Question

Q2: Marres limited has the following demand and cost functions

Demand function: P=80-3Q, where P is the unit selling price and Q is quantity in thousands

Cost function: TC=Q2+20Q+100, where TC is total cost in Ksh 000000

Required:

Optimal price to maximize profit (3mks)

Maximum profit (2mks)

Answer 1

Given that:

$TC=Q^2+20Q+100$ and

$P=80-3Q$

$TR=P×Q$

$TR=(80-3Q)Q=80Q-3Q^2$

$TR'=MR=80-6Q$

$TC'=MC=2Q+20$

$MR=MC$ at equilibrium

Then:

$80-6Q=2Q+20$

$8Q=60$

$Q^*=7.5$ in thousand

a) Optimal price to maximize profit:

$P^*=80-3(7.5)=Ksh57.5$

b) Maximum Profit, $\pi$ =$TR-TC$

At $Q^*=7.5$

$\pi=80(7.5)-3(7.5)^2-((7.5)^2+20(7.5)+100)$

$\pi=125(Ksh '000000)$