SOLUTION: p=-1/10x+150 0 ≤ x ≤ 1500 Revenue is x*p. (a) Express the revenue R as a function of x. (b) What quantity x maximizes revenue? What is the maximum revenue?

Algebra ->  Inequalities -> SOLUTION: p=-1/10x+150 0 ≤ x ≤ 1500 Revenue is x*p. (a) Express the revenue R as a function of x. (b) What quantity x maximizes revenue? What is the maximum revenue?       Log On


   

Question 974830: p=-1/10x+150
0 ≤ x ≤ 1500
Revenue is x*p.
(a) Express the revenue R as a function of x.
(b) What quantity x maximizes revenue? What is the maximum revenue?
so would i solve for x? which would be x=1500-10p
or f(p)=1500-10p
im lost and need direction
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Since Revenue = and , we can say:



The graph of this function is a parabola opening downward. Find the vertex. The -coordinate of the vertex is the value of that maximizes revenue. The -coordinate is the value of the maximum revenue.

Hint:

For a parabola with equation of the form



The -coordinate of the vertex is given by . The -coordinate of the vertex is simply the value of the function at the -coordinate of the vertex, to wit:



Just plug in the numbers and do the arithmetic.

John

My calculator said it, I believe it, that settles it