Answer
(a) cost $C = 5x^2 + 2(4)(xh) = 5x^2 + \frac{64}{x}$
(b) Domain of $C$ $(0, +∞)$
Work Step by Step
(a) If the side has length $x$ and height $h$, then $V = 8 = x^2h$, so $h = \frac{8}{x^2}$
Then the cost $C = 5x^2 + 2(4)(xh) = 5x^2 + \frac{64}{x}$
(b) The domain of $C$ is $(0, +∞)$ because $x$ can be very large (just take $h$ very small).
