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.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

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.

Approved by eNotes Editorial Team