Answer:
[tex]r\approx 1.084\ feet[/tex]
[tex]h\approx 1.084\ feet[/tex]
[tex]\displaystyle A=11.07\ ft^2[/tex]
Step-by-step explanation:
Optimizing With Derivatives
The procedure to optimize a function (find its maximum or minimum) consists in :
We know a cylinder has a volume of 4 [tex]ft^3[/tex]. The volume of a cylinder is given by
[tex]\displaystyle V=\pi r^2h[/tex]
Equating it to 4
[tex]\displaystyle \pi r^2h=4[/tex]
Let's solve for h
[tex]\displaystyle h=\frac{4}{\pi r^2}[/tex]
A cylinder with an open-top has only one circle as the shape of the lid and has a lateral area computed as a rectangle of height h and base equal to the length of a circle. Thus, the total area of the material to make the cylinder is
[tex]\displaystyle A=\pi r^2+2\pi rh[/tex]
Replacing the formula of h
[tex]\displaystyle A=\pi r^2+2\pi r \left (\frac{4}{\pi r^2}\right )[/tex]
Simplifying
[tex]\displaystyle A=\pi r^2+\frac{8}{r}[/tex]
We have the function of the area in terms of one variable. Now we compute the first derivative and equal it to zero
[tex]\displaystyle A'=2\pi r-\frac{8}{r^2}=0[/tex]
Rearranging
[tex]\displaystyle 2\pi r=\frac{8}{r^2}[/tex]
Solving for r
[tex]\displaystyle r^3=\frac{4}{\pi }[/tex]
[tex]\displaystyle r=\sqrt[3]{\frac{4}{\pi }}\approx 1.084\ feet[/tex]
Computing h
[tex]\displaystyle h=\frac{4}{\pi \ r^2}\approx 1.084\ feet[/tex]
We can see the height and the radius are of the same size. We check if the critical point is a maximum or a minimum by computing the second derivative
[tex]\displaystyle A''=2\pi+\frac{16}{r^3}[/tex]
We can see it will be always positive regardless of the value of r (assumed positive too), so the critical point is a minimum.
The minimum area is
[tex]\displaystyle A=\pi(1.084)^2+\frac{8}{1.084}[/tex]
[tex]\boxed{ A=11.07\ ft^2}[/tex]
