# Find the maximum revenue. What is the maximum monthly profit? The monthly cost and price-demand equations are C(x) = 350x + 50,000 and p = 500 – 0.025x, 0 ≤ x ≤ 20,000 The monthly cost and price-demand equations are C(x) = 350x + 50,000 and p = 500 – 0.025x, 0 ≤ x ≤ 20,000. You should use the price-demand equation to find the maximum revenue.

You need to differentiate the price demand equation with respect to x such that:

`R(x) = (500 – 0.025x)' =gt R(x) = -0.025`

You should remember that you may find the extreme values of a function solving the equation...

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You should use the price-demand equation to find the maximum revenue.

You need to differentiate the price demand equation with respect to x such that:

`R(x) = (500 – 0.025x)' =gt R(x) = -0.025`

You should remember that you may find the extreme values of a function solving the equation P'(x) = 0.

Notice that `P'(x) = -0.025 != 0` , hence, the price demand function has not a maximum value if `x in [0, 20000].`

You need to find the equation of profit such that:

`P(x) = p(x) - C(x)`

`P(x) = 500 – 0.025x - 350 x - 50000`

`P(x) = -349.975 x - 49500`

You may find the maximum profit if you solve the equation P'(x) = 0.

`P'(x) = -349.975 != 0`

Hence, evaluating the maximum profit under given conditions yields that the company has no maxim profit.

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