Word Problem. Can anyone figure this one out?

Question

. A study done for management shows that the revenue, R, in dollars, as a function of number, x, of units produced can be modeled by R(x) = 10x - 0.001x^2 while the cost, C, in dollars, can be modeled by C(x) = 5000 + 2x. What will the maximum profit be, in dollars?

Answer 1

Profit P(x) = Revenue R (x) - Cost C(x)
= 8x - 0.001x^2 - 5000

For max. profit, P'(x)=0.
=> 8 = 0.002x
=> x=4000
Ans is P(4000) = 11,000 dollars

Answer 2

Revenue R(x) = 10x - 0.001x^2
Cost C(x) = 5000 + 2x

Maximum profit is where R(x) - C(x) is maximum

Take derivative of R(x) - C(x) to get maximum (a -x^2 tells us that zero derivative will give us maximum value)

r(x) - c(x) = -0.001x^2 +8x -5000

derivative = -0.002x + 8

derivative = 0 when x = 8/0.002 = 4000

hope you learned something :-)

Answer 3

It's an easy exercise in calculus. Form a profit function from the expressions given, differentiate it with respect to x, set the derivative to zero, and solve the result. Then the resulting x will be the number of units to maximize the profit, and you can substitute this back into the profit expression to get its value.

Answer 4

profit=revenue-cost=(10x-.001x^2)-(5000+2x)
for maximum profit dr/dx must be a maximum
dr/dx=10-0.002x-2=>8-.002x,equating this to zero
x=4000 and substituting x=4000,the profit=
8(4000)-5000-16000=11000
so maximum profit for x=4000 is $11000

Answer 5

The person above is right. Taking the derivative of R - C setting the limit to 0, will find the answer.

Answer 6

isn't profic revenue - cost? write profit function and take derivative wrt x = 0 evaluate x