given the revenue function R(x) =900x-3x2, and the cost function C(x)=450x+5000, with the 0

Question

given the revenue function R(x) =900x-3x2, and the cost function C(x)=450x+5000, with the 0< x < 150 find the break even point, find the profit function, P(x), find the maximum profit, simplify, using positive exponents x-2(4x4)3/2

anonymous · Answer

The break even point is the point at which the revenue is equal to the cost, that's:
$$900x-3x^2=450x+5000 \implies 3x^2-450x+5000=0 \implies x=12.08, x=137.9$$

anonymous · Answer

The prophet is the the revenue minus the cost. So,
$$P(x)=900x-3x^2-450x-5000 \implies P(x)=450x-3x^2-5000$$

anonymous · Answer

do u think u can answer the other 2 questions

anonymous · Answer

Of course I can, but I don't have time for that. To find the maximum prophet, solve for x when the derivative equal to zero. Substitute this value of x into P(x); that would be your maximum.