A company manufacturers and sells x electric drills per month. The monthly cost and price-demand equations are
C(x)=64000+60x, p=190−x/30, 0≤x≤5000.

a) production level at max revenue = 2850
b) price to max profit = $125
c) Suppose that a 5 dollar per drill tax is imposed. Determine the number of drills that should be produced and sold in order to maximize profit under these new circumstances. ???

Question

A company manufacturers and sells x electric drills per month. The monthly cost and price-demand equations are
C(x)=64000+60x,     p=190−x/30, 0≤x≤5000.

a) production level at max revenue = 2850
b) price to max profit = $125
c) Suppose that a 5 dollar per drill tax is imposed. Determine the number of drills that should be produced and sold in order to maximize profit under these new circumstances.  ???

anonymous · Answer

how did you get the answers?

anonymous · Answer

R(x) = P(x) * x 
         = (190−x/30)*x

anonymous · Answer

to find the max revenue, take the derivative and set it equal to zero; solve for x.

anonymous · Answer

that's the number of units that give max revenue.  call it x1.
P(x1) is the price at max revenue...

anonymous · Answer

profit is R(x) -C(x) ...

anonymous · Answer

same thing, take the derivative and find the max.  ...  find the price at that production level...