A floogle company's revenue is given by the function r(x)= 200- 1600/x+8-x where R is measured in thousands of dollars.

Find the value of x which results in the maximum revenue.
x =?

Question

A floogle company's revenue is given by the function  r(x)= 200- 1600/x+8-x where R is measured in thousands of dollars.

Find the value of x which results in the maximum revenue.
x =?

campbell_st · Answer

I assume r(x) is 

$$r(x) = 200 -1600/(x+8) - x$$

then differentiating

$$r'(x) = 1600/(x+8)^2 -1$$

let r'(x) = 0 and solve for x

$$1 = 1600/(x+8)^2$$

$$(x + 8)^2 = 1600$$

$$x + 8 = \pm \sqrt{1600}$$

$$x + 8 = \pm 400$$

$$x = 392, - 408$$

sub each value into r(x) then maximum will be the larger r(x) after substiution