Please help, I've been stuck on this question for days!!

Question

For a certain company, the cost function for producing x

items is C(x)=50x+150

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 150

items.

The profit function P(x)

is the revenue function R(x)

(how much it takes in) minus the cost function C(x)

(how much it spends). In economic models, one typically assumes that a company wants to maximize its profit, or at least make a profit!

Answers to some of the questions are given below so that you can check your work.

Assuming that the company sells all that it produces, what is the profit function?

P(x)=

What is the domain of P(x) ?

Hint: Does calculating P(x)

make sense when x=−10

or x=1,000 ?

The company can choose to produce either 40

or 50

items. What is their profit for each case, and which level of production should they choose?

Profit when producing 40

items =

Number

Profit when producing 50

items =

Number

Can you explain, from our model, why the company makes less profit when producing 10 more units?

Answer 1

R(x) =-.5(x-90)^2 + 4,050

C(x) = 50x +150

0<x < 150

Profit = P(x) = R(x)-C(x) = -.5(x^2 - 180x + 8100) + 4050 -50x -150

= -.5x^2 +90x - 4050 + 4050 -50x -150

= -.5x^2 +40x -150

= -.5x^2 +40x -150

P'(x) = -x +40 =0

x = 40 = profit maximizing out put level < 150 so it's very feasible

0<40<150

max Profit = P(40)

P(40) = -.5(40)^2 -95(40) + 7950 = -800 -3800+ 7950

= 7950- 4600 = $3,350 profit

P(50) = -.5(50)^2 -95(50) + 7950 = - 1250 -4750+ 7950

= 7950 - 5000= $2,950

any output greater than or less than 40 will have less profit P(50)<P(40)

the profit function is quadratic, with leading x^2 <0

it dominates and as x gets large enough profits decline, since the x^2 term is negative, overcoming the linear x term. any x greater than 40, like 50 and profits decline

with luck there's no mistake above, but it's the basic solution method, error free or not