SL 5.7—Optimisation

Base side is $x$, height is $h$. Volume constraint: $x^{2}h=500$, so $h=\frac{500}{x^2}$.

A closed box has 2 square faces (top and bottom) plus 4 rectangular sides: $S=2x^2+4xh$.

$S(x)=2x^2+4x\left(\frac{500}{x^2}\right)=2x^2+\frac{2000}{x}$.

This matches option A exactly.

Using $500/x$ instead of $2000/x$ forgets to multiply by 4 for the four side faces.

Using $x^2$ instead of $2x^2$ forgets the box is closed (has both a top and bottom).

Using $4x^2$ incorrectly counts four square faces instead of two.

Only option A correctly accounts for 2 square faces and 4 rectangular faces.

$h^{\prime }(t)=-10t+20$.

$-10t+20=0⟹t=2$.

$t=2$ lies within $0\le t\le 4$, so it is a valid critical point.

The maximum height occurs at $t=2$ seconds.

$t=1$ comes from incorrectly setting $-10t+10=0$, a sign or coefficient slip.

$t=4$ is just the domain endpoint, not the critical point where $h^{\prime }(t)=0$.

$t=2.5$ would result from dividing 20 by 8 instead of 10, a computational error.

Only $t=2$ satisfies $h^{\prime }(t)=0$ correctly.