# Optimization of Right circular cylinder

• October 21st 2013, 10:25 PM
Jonroberts74
Optimization of Right circular cylinder
A soft drink can in the shape of a right circular cylinder is to have the capacity for 250mm^3. If the diameter of the can must be no less than 4cm and no greater than 8cm, find the dimensions of the can that will use the least amount of materials (include the top, bottom and side). What is the ratio of height:diameter of this can?

okay so I know the interval is $4\le x \le8$ where x is the diameter

Is this the right equation to use in order to minimize?

$V = \pi (\frac{d}{2})^2h$ and d and h are the two variables so to eliminate one would this be correct? changing $\frac{d}{2} \Rightarrow r$ because it seems simpler to process $V=\pi r^2\frac{250}{\pi r^2}$
• October 22nd 2013, 10:57 AM
SlipEternal
Re: Optimization of Right circular cylinder
That is the volume of a right circular cylinder, so yes. I take it you are using $h = \dfrac{250}{\pi r^2}$? But, look at what you are trying to minimize. It is not volume. You know the volume. You are trying to minimize materials used to make the can. You are not given any way of determining the volume of the aluminum to use. You are asked to calculate the materials for the top, bottom, and side of the can.
• October 22nd 2013, 11:05 AM
Jonroberts74
Re: Optimization of Right circular cylinder
I am. To continue do I input this into the surface area equation and then differentiate?

$S = 2\pi r h + 2 \pi r^2$
• October 22nd 2013, 11:14 AM
SlipEternal
Re: Optimization of Right circular cylinder
Exactly
• October 22nd 2013, 01:07 PM
Jonroberts74
Re: Optimization of Right circular cylinder
$\frac{d}{dr} [\frac{500}{r} + 2\pi r^2] \Rightarrow 4\pi r - \frac{500}{r^2}$

solving for S'(r) = 0

$4\pi r - \frac{500}{r^2}$

$\frac{4(-125+\pi r^3)}{r^2}$

$\frac{-125+ \pi r^3}{r^2}$

$-125+\pi r^3$

$\frac{\pi r^3}{125} - 1$

$x = (\frac{\sqrt[3]\pi r}{\sqrt[3]125})$ then $x^3 - 1 =0$

factor

$(x-1)(x^2 + x +1)$

$x = 1 \Rightarrow 1 = (\frac{\sqrt[3]\pi r}{\sqrt[3]125})$

so $\frac{\sqrt[3]\pi r}{5} = 1 \Rightarrow r = \frac{5}{\sqrt[3]\pi}$ is this correct?

Didn't solve $x^2 + x+1$ because it would end up with $i$ being in it
• October 22nd 2013, 01:15 PM
SlipEternal
Re: Optimization of Right circular cylinder
Here is an easier way of solving it:

$S'(r) = 4\pi r-\dfrac{500}{r^2} = 0$

$4\pi r = \dfrac{500}{r^2}$ (multiply both sides by $r^2$)

$4\pi r^3 = 500$

$r^3 = \dfrac{125}{\pi}$

$r = \dfrac{5}{\sqrt[3]{\pi}}$

Anyway, that is the correct answer for $r$, but your original problem asks you to use the closed interval method. So, try:

$S(2), S(4), \mbox{ and } S\left(\dfrac{5}{\sqrt[3]{\pi}}\right)$
• October 22nd 2013, 01:28 PM
Jonroberts74
Re: Optimization of Right circular cylinder
would I solve for those values of r then multiply them by 2 because it wants diameter?
• October 22nd 2013, 01:37 PM
SlipEternal
Re: Optimization of Right circular cylinder
You will do $\dfrac{125}{\pi r^3} = \dfrac{h}{d}$ (which is the ratio height to diameter the problem wants)
• October 22nd 2013, 02:06 PM
Jonroberts74
Re: Optimization of Right circular cylinder
I need the diameter and height as well. I solved for the values of r getting the smallest amount with $r = \frac{5}{\sqrt[3]\pi}$ for approximately 119.85

then I took that value of r and times it by two for a diameter of 6.83

solving the h/d ratio

I got 125/125 or 1:1 which means both h and d are 6.83 approximately

also subbed the value of r back into the formula that gave me the value of h which gave me h = 6.83, and r=6.83 so that is 1:1 as well

But, where did $\frac{125}{\pi r^3} = \frac{h}{d}$ come form?
• October 22nd 2013, 03:17 PM
SlipEternal
Re: Optimization of Right circular cylinder
$h = \dfrac{250}{\pi r^2}, d = 2r, \dfrac{h}{d} = \dfrac{\left(\dfrac{250}{\pi r^2}\right)}{2r} = \dfrac{125}{\pi r^3}$