## Precalculus (6th Edition) Blitzer

$\dfrac{x^3+16}{x}$
Let $y$ be the height of the box. Surface Area, $A=x^2+4xy$ $V=x^2y=4 \implies y=\dfrac{V}{x^2}=\dfrac{4}{x^2}$ Surface Area, $A=x^2+4xy=x^2+4x \times \dfrac{4}{x^2}$ or, $A=x^2+\dfrac{16}{x}=\dfrac{x^3+16}{x}$