The profit is a max (or min) if `(dP)/(dx)=0`.

`(dP)/(dx)=8-0.04x`

`=0`

when

`x=8/0.04=200`

Is it a maximum?

`(d^2 P)/(dx^2) = -0.04 < 0` for all x, so we have a maximum.

When `x = 200`, `P = $800`.

So if the company refines `200` barrels per day, the maximum profit of `$800` is reached.

1002003004002004006008001000xPOpen image in a new page

Graph of `P=8x-0.02x^2`.

The maximulm point, `(200, 800)` is indicated on the graph with a magenta dot.