Question

A metal cylinder can with an open top and closed bottom is to have volume 4 cubic feet. Approximate the dimensions that require the least amount of material. What would this amount be?

Answer 1

Answer:

$r\approx 1.084\ feet$

$h\approx 1.084\ feet$

$\displaystyle A=11.07\ ft^2$

Step-by-step explanation:

Optimizing With Derivatives

The procedure to optimize a function (find its maximum or minimum) consists in :

•  Produce a function which depends on only one variable
•  Compute the first derivative and set it equal to 0
•  Find the values for the variable, called critical points
•  Compute the second derivative
•  Evaluate the second derivative in the critical points. If it results positive, the critical point is a minimum, if it's negative, the critical point is a maximum

We know a cylinder has a volume of 4 $ft^3$. The volume of a cylinder is given by

$\displaystyle V=\pi r^2h$

Equating it to 4

$\displaystyle \pi r^2h=4$

Let's solve for h

$\displaystyle h=\frac{4}{\pi r^2}$

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

$\displaystyle A=\pi r^2+2\pi rh$

Replacing the formula of h

$\displaystyle A=\pi r^2+2\pi r \left (\frac{4}{\pi r^2}\right )$

Simplifying

$\displaystyle A=\pi r^2+\frac{8}{r}$

We have the function of the area in terms of one variable. Now we compute the first derivative and equal it to zero

$\displaystyle A'=2\pi r-\frac{8}{r^2}=0$

Rearranging

$\displaystyle 2\pi r=\frac{8}{r^2}$

Solving for r

$\displaystyle r^3=\frac{4}{\pi }$

$\displaystyle r=\sqrt[3]{\frac{4}{\pi }}\approx 1.084\ feet$

Computing h

$\displaystyle h=\frac{4}{\pi \ r^2}\approx 1.084\ feet$

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

$\displaystyle A''=2\pi+\frac{16}{r^3}$

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

$\displaystyle A=\pi(1.084)^2+\frac{8}{1.084}$

$\boxed{ A=11.07\ ft^2}$