Production Levels and Pricing

Question

 A company manufacturers and sells x electric drills per month. The monthly cost and price-demand equations are

C(x)=73000+70x,

p=220−x30,0≤x≤5000.

Find the production level that results in the maximum profit.

 Find the price that the company should charge for each drill in order to maximize profit.

Answer 1

We want to maximize the profit, so we will first need to find the profit function (Profit = Revenue - Cost), and take the derivative of that function to get P'(x). We set the derivative P'(x) to zero to find where the slope is zero to find any critical points, and then verify which one will give us a maximum point of profit (not a minimum).

P(x) = R(x) - C(x)

We need to get our Revenue function using R(x) = xp (x number of units sold times the price per unit - the demand function)

R(x) = x p(x)

R(x) = x ( 220 - x/30 ) = 220x - x²/30

P(x) = R(x) - C(x) = ( 220x - x²/30 ) - ( 73000 + 70x) = -(1/30) x² + 150x - 73000

P'(x) = -(1/15) x + 150

***Because P'(x) tells us the slope at any x we can exploit the derivative function by setting the slope to be zero and then finding the x values at which the slope is zero.

P'(x) = -(1/15) x + 150

0 = -(1/15) x + 150

(1/15) x = 150

x = 2250

x = 2250 is now a "possible" Max or Min. We need to find out which one it is by either using P'(x) to check the slope on either side of it or find P''(x) to check the concavity at x = 2250. It is simple to find P''(x) = -(1/15) which means it is always negative or always concave down. Because P(x) is always concave down, it means that x=2250 is a Max point where the slope is zero for an instant.

So, the maximum profit occurs when the company produces 2250 drills.

The price they charge comes from the demand function (price per unit):

p(x) = 220 - x/30

p(2250) = 220 - 2250/30 = $145