You find q by taking the derivatives of both R(q) and C(q) and setting the derivatives equal to each other.
R'(q)=750
C'(q)=10q
q=75
Total profit is $22125, once you plug it in.
I don't understand how to get q for this problem? The next steps for the problem, you just need to plug q in, but I'm not sure what to do?
Do you set revenues equal to expenses?
Problem:
Revenue is given by
R(q) = 750q
and cost is given by
C(q) = 6000 + 5q2.
1) At what quantity is profit maximized?
q =
2) What is the total profit at this production level?
For problem 2, it's clear you just need to plug q in. I just don't know how to get q.
Please help?
2 answers
profit is revenue - cost
P(q) = R(q)-C(q)
= 750q - (6000+5q^2)
= -5q^2 + 750q - 6000
dP/dq = -10q + 750
So, P is maximuzed when q=75
revenue = cost is the break-even point. Not much profit there...
P(q) = R(q)-C(q)
= 750q - (6000+5q^2)
= -5q^2 + 750q - 6000
dP/dq = -10q + 750
So, P is maximuzed when q=75
revenue = cost is the break-even point. Not much profit there...