# Thread: Substition Within an Optimization Problem

1. ## Substition Within an Optimization Problem

I am working on an optimization problem right now and ran into the following situation.

$\displaystyle V = piR^2H = 8$
$\displaystyle H = 8/piR^2$
$\displaystyle H = 2srt(2)/piR$

$\displaystyle A = H2piR+piR^2$
$\displaystyle A = (2rt(2)/piR)2piR+piR^2$
$\displaystyle A = 4srt(2)+piR^2$

$\displaystyle dA/dR = 2piR$

I think I have done this wrong, but I am fairly tired right now and can't seem to find my mistake.

2. Originally Posted by Vanilla
I am working on an optimization problem right now and ran into the following situation.

$\displaystyle V = piR^2H = 8$
$\displaystyle H = 8/piR^2$
$\displaystyle H = 2srt(2)/piR$ incorrect

$\displaystyle A = H2piR+piR^2$
$\displaystyle A = (2rt(2)/piR)2piR+piR^2$
$\displaystyle A = 4srt(2)+piR^2$

$\displaystyle dA/dR = 2piR$

I think I have done this wrong, but I am fairly tired right now and can't seem to find my mistake.
Are you trying to minimise the surface area of the cylinder for a fixed volume of 8 cubic units?

3. Not sure where I got the idea that $\displaystyle piR^2 = (piR)^2$.

Are you trying to maximise the surface area of the cylinder for a fixed volume of 8 cubic units?
Trying to minimize the surface area. However; I know how to solve the rest of the problem. Thanks for the help.

Edit: Actually, I got stuck again.

$\displaystyle A = H2piR + piR^2$ $\displaystyle H = 8/piR^2$
$\displaystyle A = (8/piR^2)*2piR + piR^2$
$\displaystyle A = (16piR/piR^2) + piR^2$
$\displaystyle A = (16/R) + piR^2$
$\displaystyle dA/dR = (R - 16/R^2) + 2piR$

Then solving for dA/dR = 0.

$\displaystyle 0 = (R - 16/R^2) + 2piR$
$\displaystyle -2piR = (R - 16)/R^2$
$\displaystyle -2piR^3 = R - 16$

And from there I can't find a way to isolate R, probably because I made some silly algebraic error.

4. Hi Vanilla,

The volume of a cylinder is $\displaystyle {\pi}R^2H$

The curved surface area is $\displaystyle 2{\pi}RH$

The flat surface area is $\displaystyle {\pi}R^2$

The total surface area of a cylinder closed at both ends (surface area of a solid cylinder) is

$\displaystyle 2{\pi}RH+2{\pi}R^2$

$\displaystyle V={\pi}R^2H=8\Rightarrow\ H=\displaystyle\frac{8}{{\pi}R^2}$

Substitute the equation for H into A to get an equation for SA in terms of the single variable R...

Surface area $\displaystyle =\displaystyle\ 2{\pi}R\ \frac{8}{{\pi}R^2}+2{\pi}R^2=\frac{16}{R}+2{\pi}R^ 2$

Differentiating with respect to R and setting derivative to zero
finds the value of R that gives minimum surface area.

Double check whether or not the cylinder is closed.

5. The cylinder is open on one end. So the $\displaystyle 2{\pi}R^2$ in the surface area equation would instead be $\displaystyle {\pi}R^2$, correct?
So differentiating $\displaystyle \frac{16}{R}+{\pi}R^2$ would still leave me with $\displaystyle \frac{R-16}{R^2}+2{\pi}R$ for which I can't seem to isolate R when solving for 0.

6. Yes,

you only need to adjust for that before calculating the derivative.

7. Double post.

8. I think I got it.
$\displaystyle \frac{16}{R}+{\pi}R^2$ can be simpified to $\displaystyle \frac{16+{\pi}R^3}{R}$, which then differentiates to $\displaystyle \frac{2{\pi}R^3-16}{R^2}$. And for $\displaystyle \frac{dA}{dR} = 0$, $\displaystyle R = (\frac{8}{\pi})^\frac{1}{3}$. Thanks for the help!

9. Originally Posted by Vanilla
I think I got it.
$\displaystyle \frac{16}{R}+{\pi}R^2$ can be simpified to $\displaystyle \frac{16+{\pi}R^3}{R}$, which then differentiates to $\displaystyle \frac{2{\pi}R^3-16}{R^2}$. And for $\displaystyle \frac{dA}{dR} = 0$, $\displaystyle R = (\frac{8}{\pi})^\frac{1}{2}$. Thanks for the help!
Yes,

but as the numerator is zero for R giving minimum area, you need to take the cube root
rather than the square root.