Question

Jane manages a small bakery that bakes only "make-to order" cakes. Her total cost (TC) and total revenue (TR) functions are as follows:

 

 

TC = 50 + 80Q - 10Q² + Q³

TR = 100Q - 8Q²

 

The bakery has a capacity of handling 30 units at any one time.

i. Determine the price she should charge if she wants to maximise total revenue.

ii. Compute the price she should charge if she wants to maximise profits.

Answer 1

Solution

Total Cost:

$TC=50+80Q-10Q^2+Q^3$ (1)

Total Revenue:

$TR=100Q-8Q^2$ (2)

(i)

The price that maximizes Total Revenue is attained when the derivative of TR with respect to Q is 0

Derivative of TR with respective to Q

$dTR/dQ=100-16Q=0$

$16Q=100$

$Q=6.25$

$Q≈6$ units (3)

Substitute the value of Q into (2) to obtain maximum Total Revenue

$TR=100(6)-8(6)^2=600-288=312$

$TR=$ $312$

Price(P) at $Q=6$ is given by:

$P=TR/Q=312/6$

$P=52$

So, the price that maximizes the Total Revenue will be 52

(ii)

The price that will maximize profits will be attained when the derivative of the profit function with respect to Q is 0

$π(Q)≈TR-TC$

$π(Q)≈(100Q-8Q^2)-(50+80Q-10Q^2+Q^3)$

$π(Q)=20Q+2Q^2-Q^3-50$ (4)

Derivative of π(Q) with respect to Q

$dπ(Q)/dQ=20+4Q-3Q^2=0$

$Q=(-4±√((4)^2-4(20(-3)) ))/2(-3) =(-4±√256)/(-6)$

$Q=(-4±16)/(-6)$

$Q=-2 or3.33$

Substitute the value of Q, 3 into (2) to Obtain Total Revenue at Q=3

$TR=100(3)-8(3)^2=300-72=228$

Price(P) at $Q=3$ is given by:

$P=TR/Q=288/3$

$P=96$

So, the price that will maximize profit is 96