Cheapest can problem A cylindrical can which must hold 1000 mL is set to be designed so the least amount of material is necessary to make the can.


*

*What should the radius be?

*What is the height of the can?

*What is the minimum surface area?


I'm not really sure how or where to start...please help!
Thanks,
Bill
 A: The surface area $A$ of a cylinder is given by:
$$A = 2\pi r^2 + h\times 2\pi r\tag{1}$$
Where $r$ is the radius of the cylinder, and $h$ is its height.
Volume, V, is given by 
$$V = \pi r^2 \times h = 1000\tag{2}$$
We want to minimize surface area that yields a can of volume $1000$. That will minimize the cost of the material to build the can. 
Express height $h$ as a function of $r$, using equation $(2)$, 
$$ h=\frac{1000}{\pi r^2} \tag{3}$$
...Then substitute this expression for $h$ into the formula of $(1)$. This will give us surface area $A$ as a function of one variable, $r$:
$$ A(r) = 2\pi r^2 + 2\pi r \left(\frac{1000}{\pi r^2}\right) = 2\pi r^2 + \frac{2000}{r}\tag{A(r)}$$
We can find the value of $r$ by finding where $A(r)' = 0$:
So take the derivative of this last equation with respect to $r$, set $A'(r)$ equal to $0$ to solve for the only possible minimal value of $r$:
$$A(r)' = 4\pi r - \frac{2000}{r^2}$$
$$A(r)' = 0 \iff 4\pi r - \frac{2000}{r^2} = 0 \iff 4\pi r^3 - 2000 = 0 $$
$$\iff r^3 = \frac{2000}{4\pi} \iff r = \sqrt[\large 3]{\frac{2000}{4\pi}} \iff r = \sqrt[\large 3]{\frac{500}{pi}}$$
N.B. To confirm that $r = \sqrt[3]{\frac{500}{\pi}}$ gives the minimum surface area $(1)$, evaluate $A(r)$ at $r\lt \sqrt[3]{\frac{500}{\pi}}$, and $r >  \sqrt[\large 3]{\frac{500}{\pi}}$. We want to ensure that those values are greater than $A(\sqrt[\large 3]{\frac{500}{\pi}})$. Then you know for sure that $r = \sqrt[\large 3]{\frac{500}{\pi}}$ will indeed give the minimum surface area.
Now we solve for $h$ by substituting $r = \sqrt[\large 3]{\frac{500}{\pi}}$ into our equation for height given by $(3)$.
Then compute $A$ as given in $(1)$ using these values of $r$, $h$.

If you'd like to check your computations:
$$r = \sqrt[\large 3]{\frac {500}{\pi}}\approx 5.419$$
$$ h=\frac{1000}{\pi r^2} = \frac{1000}{\pi^{1/3}500^{2/3}}\approx 10.84$$
$$ A_{\text{minimum}} = 2\pi r^2 + h\times 2\pi r = 2\pi r(r + h) \approx 544.401$$
A: Hints:
Let $r=$ the radius of the cylinder and $h$ = the height of the can.
Then an equation for the surface area is:
$$
SA=2\pi r^2 + 2\pi r h
$$
We also have a constraint on the volume:
$$V=\pi r^2h=1000
$$
We can minimize the $SA$ equation by taking the derivative, setting it equal to 0, and then using the Volume equation as a constraint. This will build the cheapest can.
Continuing on, we have:
$$
h=\frac{1000}{\pi r^2}\\
SA=2\pi r^2 + 2\pi r \left(\frac{1000}{\pi r^2}\right)=2\pi r^2 + \frac{2000}{r}
$$
Now, take the derivative of this last equation with respect to $r$, set the result equal to $0$ and solve for the minimal value of $r$:
$$
SA(r)'=4\pi r -\frac{2000}{r^2}=0
$$
