Please help me solve this problem I have been struggling wit

Please help me solve this problem, I have been struggling with it for hours!

Your college newspaper, The Collegiate Investigator, sells for 90¢ per copy. The cost of producing x copies of an edition is given by

C(x) = 10 + 0.10x + 0.001x² dollars.

(a) Calculate the marginal revenue R\'(x) and profit P\'(x) functions. HINT [See Example 2.]

(b) Compute the revenue and profit, and also the marginal revenue and profit, if you have produced and sold 500 copies of the latest edition.

Interpret the results.

The approximate  ---Select--- profit loss from the sale of the 501st copy is $  .

(c) For which value of x is the marginal profit zero?
x =  copies

Interpret your answer.

The graph of the profit function is a parabola with a vertex at x =  , so the profit is at a maximum when you produce and sell  copies.

Solution

a ) for x quantity of product
Revenue = price times quantity
R(x) = 0.9x
cost of product
C(x) = 50 + 0.1x + 0.001x²
profit
P(x) = R(x) - C(x) = 0.9x - (10 + 0.1x + 0.001x²)
P(x) = -10 + 0.8x - 0.001x²
marginal revenue
R\'(x) = dR(x)/dx = 0.9 dollars
>> marginal profit
P\'(x) = dP(x)/dx = (0.8 - 0.002x) dollar
b) x = 500, then
R(x) = 0.9*500 = 450 dollars
P(x) = -10 + 0.8*500 - 0.001*500² = 140 dollars
R\'(x) = 0.9 dollars
P\'(x) = 0.8 - 0.002*500 = -0.2 dollars

if x = 501 ---> R(x) = 0.9*501 = 450.9 dollar

c )
P\'(x) = 0
0.8 - 0.002x = 0
x = 0.8/0.002
x = 400 copies

the max profit reach when P\'(x) = 0 or x = 400 copies and will give of
P(x) = -10 + 0.8*400 - 0.001*400²
P(x) = 150 dollars

y = ax^2 + bx + c is the parabola
y = -0.001x^2+0.8x+10
x-coordinate of vertex is -b/2a
= -(0.8) / (2)(-0.001)
= -0.8 / (2)(-0.001) = 400
The graph of the profit function is a parabola with a vertex at x =400

y = -0.001(400)^2+0.8(400)+10
= -160+320+10 = 170

So the profit is at a maximum when you produce and sell 400 copies.