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anonymous
 5 years ago
given the revenue function R(x) =900x3x2, 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 x2(4x4)3/2
anonymous
 5 years ago
given the revenue function R(x) =900x3x2, 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 x2(4x4)3/2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The break even point is the point at which the revenue is equal to the cost, that's: \[900x3x^2=450x+5000 \implies 3x^2450x+5000=0 \implies x=12.08, x=137.9\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The prophet is the the revenue minus the cost. So, \[P(x)=900x3x^2450x5000 \implies P(x)=450x3x^25000\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do u think u can answer the other 2 questions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Of 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.
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