Glass optimization

The area of the glass as a cylinder is $\pi { r }^{ 2 }+2\pi hr=40\pi$ . Keep in mind that the glass is open at the top (Technically speaking, the area is twice this number because of the inner surface, but it should be fairly obvious from the wording regarding the sheet that this is a non-issue).

Expressing for $h=\frac { 40-{ r }^{ 2 } }{ 2r }$

Now we recall the volume formula: $V=\pi { r }^{ 2 }h$ substituting h and calculating yields

$V=20\pi r-\frac { \pi { r }^{ 3 } }{ 2 }$

Differentiation and setting the value for 0 will allow us to obtain the maximum value for r

$\frac { dV }{ dr } =20\pi -\frac { 3\pi { r }^{ 2 } }{ 2 } =0$

which yields

$r=\sqrt { \frac { 40 }{ 3 } }$

plugging this value into $h=\frac { 40-{ r }^{ 2 } }{ 2r }$ shows that h equals r. This can be done via calculator and noticing the decimals or more elegant, by dividing the expression $h=\frac { 40-{ r }^{ 2 } }{ 2r }$ with $r$ on both sides and plugging in $r^{2}$ to show that $\frac { h }{ r } =1$

It follows that $\frac { h }{ r } =1$