For a certain company, the cost function for producing x items is C(x)=50x+200 and the revenue function for selling x items is R(x)=−0.5(x−90)2+4,050. The maximum capacity of the company is 130 items.

Question

For a certain company, the cost function for producing x items is C(x)=50x+200 and the revenue function for selling x items is R(x)=−0.5(x−90)2+4,050. The maximum capacity of the company is 130 items.

Answer 1

Profit = R-C = -(x-90)^2/2 +4,050 - 50x -200 =

-(x^2 -180x +8100)/2 -50x +3,850 =

-x^2/2 +90x -4050 +4500 -50x +3,850 =

-x^/2+40x -200

Maximum profit generating output is found by taking the derivative and setting it equal to zero

R-C = -x^2/2 + 40x -200

R'-C' = -x + 40 =0

x = 40,

Maximum profit is found by substituting x=40 into the original profit equation, R-C

-(1/2)(40)^2 +40(40) -200 = -800 + 1600 -200 = $600 profit

Answer 2

I interpreted the question to be:

"For a certain company, the cost function for producing x items is C(x)=50x+200 and the revenue function for selling x items is R(x)=−0.5(x−90)^2+4,050. The maximum capacity of the company is 130 items. Find the number of items to sell in order to maximize the profit (defined as revenue minus cost)."

The function that expresses the profit is P(x) = R(x) - C(x) = -0.5 (x-90)^2 + 4050 - (50x + 200). If one allows for selling of non-integer amounts of items, then we can find the maximum profit for x in [0,130] by finding the critical numbers for P(x).

We do this by finding the derivative P'(x) = -0.5 * 2 (x-90) - (50) = -x + 90 - 50 = 40 - x.

Critical numbers of P(x) are x-values in the domain of P where P'(x) = 0 or where P'(x) is undefined. For x in [0,130], P'(x) is always defined, so the only critical numbers will be where P'(x) = 0.

Setting P'(x) = 0 yields 40 - x = 0, or simply x = 40 units.

Since P is continuous on [0,130], the Extreme Value Theorem guarantees that the maximum and minimum values on P will be at the endpoints of [0,130] or at the critical numbers.

So let's check x₁=0, x₂=40 and x₃=130:

P(0) = -0.5 (0 - 90)^2 + 4050 - (50*0 + 200) = -200 dollars

P(40) = -0.5 (40 - 90)^2 + 4050 - (50*40 + 200) = 600 dollars

P(130) = -0.5 (130 - 90)^2 + 4050 - (50*130 + 200) = -3450 dollars

Therefore, in order to maximize profit, we would want to sell 40 units.