Asked by Anonymous
The profit function for a computer company is given by P(x)=−x^2+36x−26 where x is the number of units produced (in thousands) and the profit is in thousand of dollars.
a) Determine how many (thousands of) units must be produced to yield maximum profit. Determine the maximum profit.
(thousands of) units =
maximum profit = thousand dollars
b) Determine how many units should be produced for a profit of at least 40 thousand.
more than (thousands of) units
Thanks so much!
less than (thousands of) units
a) Determine how many (thousands of) units must be produced to yield maximum profit. Determine the maximum profit.
(thousands of) units =
maximum profit = thousand dollars
b) Determine how many units should be produced for a profit of at least 40 thousand.
more than (thousands of) units
Thanks so much!
less than (thousands of) units
Answers
Answered by
Reiny
take the derivative, set it equal to zero and solve for x
Plug that x back into P(x) to get your maximum profit.
b) −x^2+36x−26 ≥ 40
x^2 - 36x + 66 ≤ 0
let's use the = sign
x^2 - 36x + 66 = 0
use the quadratic formula, you will get two positive x's
lower ≤ x ≤ larger
verify your answer by looking at this graph
http://www.wolframalpha.com/input/?i=plot+y+%3D+%E2%88%92x%5E2%2B36x%E2%88%9226
Plug that x back into P(x) to get your maximum profit.
b) −x^2+36x−26 ≥ 40
x^2 - 36x + 66 ≤ 0
let's use the = sign
x^2 - 36x + 66 = 0
use the quadratic formula, you will get two positive x's
lower ≤ x ≤ larger
verify your answer by looking at this graph
http://www.wolframalpha.com/input/?i=plot+y+%3D+%E2%88%92x%5E2%2B36x%E2%88%9226
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