# Math Help - Need help with an optimization problem

1. ## Need help with an optimization problem

hi, i have a question that goes,

"a cylindrical tin of particular volume is to be made using as little material as possible. find the ratio of the height to the radius(the tin is closed both ends)"

""a cylindrical tin of particular volume" --> volume is constant.
"is to be made using as little material as possible" --> minimize surface area
"find the ratio of the height to the radius" --> solve for h/r

Write the formulas for the volume and surface areas.
V = pi r^2 h
A = 2 pi r^2 + 2 pi rh

rewrite A in terms of a single variable. Remember you can treat V as a constant.
h = V / (pi r^2)
A = 2 pi r^2 + 2 V / r

Minimize
4 pi r - 2V/r^2 = 0
r^3 = V/(2 pi)

Solve for the desired value
h/r = V / (pi r^3) = V / (V/2) = 2"

but its this bit im having a hard time understanding below.

"Minimize
4 pi r - 2V/r^2 = 0
r^3 = V/(2 pi)

Solve for the desired value
h/r = V / (pi r^3) = V / (V/2) = 2"

someone please explain it to me step by step

Thank you

2. Originally Posted by nomanslan
hi, i have a question that goes,

"a cylindrical tin of particular volume is to be made using as little material as possible. find the ratio of the height to the radius(the tin is closed both ends)"

""a cylindrical tin of particular volume" --> volume is constant.
"is to be made using as little material as possible" --> minimize surface area
"find the ratio of the height to the radius" --> solve for h/r

Write the formulas for the volume and surface areas.
V = pi r^2 h
A = 2 pi r^2 + 2 pi rh

rewrite A in terms of a single variable. Remember you can treat V as a constant.
h = V / (pi r^2)
A = 2 pi r^2 + 2 V / r
Just take the derivative of A and set it equal to 0.
$A' = 2 \pi (2r) - \frac{2V}{r^2}$

-Dan

3. but why does r^3 = V/(2 pi)?

4. Originally Posted by nomanslan
but why does r^3 = V/(2 pi)?
Just take the derivative of A and set it equal to 0.
$A' = 2 \pi (2r) - \frac{2V}{r^2}$

Setting it equal to 0 as I suggested
$2 \pi (2r) - \frac{2V}{r^2} = 0$

Multiply both sides by $r^2$:
$4 \pi r^3 - 2V = 0$

$4 \pi r^3 = 2V$

Divide both sides by $4 \pi$:
$\displaystyle r^3 = \frac{2V}{4 \pi} = \frac{V}{2 \pi}$

Since you seem to be missing the logic of the problem, let me remind you that, since you are minimizing the area, this value of r will be the value of r that maximizes or minimizes A.

'-Dan