My answers are not matching those in the back of the book so I would like to try and get some feedback to see where I went wrong.

**14)** The equation for the least material used would be $A=h(2\pi r)+8\pi r^2$, or (height)(diameter of the top / bottom) + (surface area of the top + bottom). Since the volume = 1, I made the secondary equation $1=h \pi r^2$ which can be solved for h: $h=\frac{1}{\pi r^2}$. Plugging that back into the original surface area equation, I got: $A=\frac{2\pi r}{\pi r^2}+8r^2$ or just $A=8r^2+\frac{2}{r}$.

Taking the derivative of that, I got $A'=16r-\frac{2}{r^2}$ or $\frac{16r^3-2}{r^2}$.

Set that equal to 0: $0=16r^3-2$ (the $r^2$ in the denom. is multiplied out).

Do some algebra and $r=\frac{1}{2}$.

I think the feasible domain would just be $[0,\infty]$, and the 1st derivative test shows that there's a minimum at $r=\frac{1}{2}$.

The area of the top or bottom of this cylinder would be $\pi\frac{1}{2}^2$ or $\frac{\pi}{4}$. The height would be $\frac{1}{\pi\frac{1}{2}^2}$ or $\frac{4}{\pi}$.

The ratio of height to radius would be $\frac{\frac{4}{\pi}}{\frac{1}{2}}$ or $\frac{8}{\pi}$ (which is the only part that I got right, implying that I did the calculations right for the wrong formula? or something).

**19)** With x being the length of the side of the square you're cutting out, the equation for the volume of the box would be $V=x(1-2x)(\frac{1}{2}-2x)$ or $\frac{8x^3-6x^2+x}{2}$.

The derivative of this would be $\frac{(24x^2-12x+1)(2)}{4}$.

Equal to 0: $0=24x^2-12x+1$.

$x=.3943, x=.1057$

Feasible domain: $[0,\frac{1}{4}]$, get rid of the first solution for x.

1st der. test, x is a local max.

$h=.1057$

$w=.2886$

$l=.7886$