Answer
(A) The minimum cost occurs at $r \approx3.4 cm$ and $h\approx13.8 cm$
(B) It will be taller than that in option (A)
(C) $r \approx3.2 cm$, $h\approx16.0 cm$ and $c\approx4.76 cents$
Work Step by Step
(A) The volume of a cylinder is: $V={\pi}r^{2}h$ =$500cm^{3}$
We can make the height 'h' the subject of the formula as: $h=\frac{V}{{\pi}r^{2}}$
Furthermore, the total surface area of a cylinder is: $S=2{\pi}r^{2}+2{\pi}rh$
Using the data provided, "the top/bottom costs $0.02 \frac{cents}{cm^{2}}$ while the material for the sides costs $0.01 \frac{cents}{cm^{2}}$." Thus:
$c=2{\pi}r^{2}\times0.02+2{\pi}rh\times0.01=$$(0.04{\pi})r^{2}+\frac{2V}{r}\times0.01$
So, $c=(0.04{\pi})r^{2}+\frac{10}{r}$
Then, we plot the graph of cost (c) versus the radius (r) for different values of 'r' as shown in plot 1 of the attached figure. It can be inferred from the figure that the minimum cost occurs t $r \approx3.4 cm$, and height (h):
$h=\frac{V}{{\pi}r^{2}}$ $=\frac{500}{{\pi}\times3.4^{2}}=13.8 cm$
(B) Since the cost is a function of the lateral sides and two times the surface area of a square of side 2r. The surface area of the two squares will be more than that of two circles with radius r. Thus, we can conclude that it will be taller than that in option (A)
(C) In this case, the cost: $c=2(Area) \times0.02+2{\pi}rh\times0.01=$
Area of a square of side (2r): $ A=2r\times2r=4r^{2}$, then:
$c=2(4r^{2})\times0.02+2{\pi}rh\times0.01=0.16r^{2}+\frac{10}{r}$
Also, if we plot the graph of cost versus radius as shown in plot 2 of the attached file, the minimum cost occurs at t $r \approx3.2 cm$, where $c\approx 4.76 cents$ and height (h):
$h=\frac{V}{{\pi}r^{2}}$ $=\frac{500}{{\pi}\times3.2^{2}}\approx16 cm$
The result corresponds with the observation in option (B).
