Answer:
[tex]r=4\ cm,\ h=4\ cm[/tex]
Step-by-step explanation:
Minimization
Optimization is the procedure leading to find the values of some parameters that maximize or minimize a given objective function. The parameters could have equality and inequality restrictions. If only equality restrictions hold, then we can use the derivatives to find the possible maximum or minimum values of the objective function.
The problem states we need to minimize the amount of material needed to manufacture the cylindrical can. The material is the surface area of the can. If the can has height h and radius r on the base, then the surface area is
[tex]A=2\pi rh+\pi r^2[/tex]
Note there is only one lid at the bottom (open at the top), that is why we added only the surface area of one circle.
That is our objective function, but it's expressed in two variables. We must find a relation between them to express the area in one variable. That is why we'll use the given volume (We'll assume the volume to be [tex]64\pi cm^3[/tex] because the question skipped that information).
The volume of a cylinder is
[tex]V=\pi r^2h[/tex]
We can solve it for h and replace the formula into the formula for the area:
[tex]\displaystyle h=\frac{V}{\pi r^2}[/tex]
Substituting into the area
[tex]\displaystyle A=2\pi r\cdot \frac{V}{\pi r^2}+\pi r^2[/tex]
Simplifying
[tex]\displaystyle A=\frac{2V}{ r}+\pi r^2[/tex]
Now we take the derivative
[tex]\displaystyle A'=-\frac{2V}{ r^2}+2\pi r[/tex]
Equating to 0
[tex]\displaystyle \frac{-2V+2\pi r^3}{ r^2}=0[/tex]
Since r cannot be 0:
[tex]-2V+2\pi r^3=0[/tex]
[tex]\displaystyle r=\sqrt[3]{\frac{V}{\pi}}[/tex]
Since [tex]V=64\pi[/tex]
[tex]\displaystyle r=\sqrt[3]{\frac{64\pi}{\pi}}=4[/tex]
[tex]r=4\ cm[/tex]
And
[tex]\displaystyle h=\frac{64\pi}{\pi 4^2}=4[/tex]
[tex]h=4\ cm[/tex]
Summarizing:
[tex]\boxed{r=4\ cm,\ h=4\ cm}[/tex]
