Rectangular box without a top has a volume of 216 in$^3$. Find dimensions of the box with the smallest surface area. Use second derivative test.
So this is what I came up with not sure where it will lead me. $$ 216=xyz\\ z=216/xy\\ $$ So $$ \text{Surface Area}= 2xy+2xz+yz = 2xy+2x(216/xy)+y(216/xy). $$ That's all I got, don't know where to got from there.
We don't need any calculus to find the minimum surface area - AM-GM works fine.
Solution 1. AM-GM
We have the surface area as $xy+2yz+2zx$ with the constraint of $xyz=216$.$$xy+2yz+2zx = xy+\frac{432}{x}+\frac{432}{y} \ge 3 \sqrt[3]{xy \cdot \frac{432}{x} \cdot \frac{432}{y}} = 3\sqrt[3]{432^2} = 108 \sqrt[3]{4}$$ The equality holds at $x=y=6\sqrt[3]{2}$ and $z=3\sqrt[3]{2}$.
Solution 2. Second Derivative Test
Again, the surface area is $f(x,y)=xy+\frac{432}{x}+\frac{432}{y}$.
Now $f_x(x,y)=y-\frac{432}{x^2}=0$ and $f_y(x,y)=x-\frac{432}{y^2}=0$.
This gives $x^2y=xy^2=432$, so $x=y=6\sqrt[3]{2}$, and $z=3\sqrt[3]{2}$ follows.
Now $f_{xx}(x,y)=\frac{864}{x^3}$, $f_{yy}(x,y)=\frac{864}{y^3}$, and $f_{xy}(x,y)=f_{yx}(x,y)=1$.
We have $$D = f_{xx}(x,y) \cdot f_{yy}(x,y) - f_{xy}^2(x,y) = \frac{864^2}{x^3y^3}-1 = \frac{864^2}{432^2}-1=3 > 0$$
So we have a local minimum. We are done.