# Math Help - 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.

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

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

$dA/dR = 2piR$

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.

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

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

$dA/dR = 2piR$

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 [LaTeX ERROR: Convert failed] .

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.

[LaTeX ERROR: Convert failed] [LaTeX ERROR: Convert failed]
[LaTeX ERROR: Convert failed]
[LaTeX ERROR: Convert failed]
[LaTeX ERROR: Convert failed]
[LaTeX ERROR: Convert failed]

Then solving for dA/dR = 0.

[LaTeX ERROR: Convert failed]
[LaTeX ERROR: Convert failed]
[LaTeX ERROR: Convert failed]

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 ${\pi}R^2H$

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

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

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

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

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

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

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

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 [LaTeX ERROR: Convert failed] in the surface area equation would instead be ${\pi}R^2$, correct?
So differentiating [LaTeX ERROR: Convert failed] would still leave me with [LaTeX ERROR: Convert failed] 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.
[LaTeX ERROR: Convert failed] can be simpified to [LaTeX ERROR: Convert failed] , which then differentiates to [LaTeX ERROR: Convert failed] . And for [LaTeX ERROR: Convert failed] , [LaTeX ERROR: Convert failed] . Thanks for the help!

9. Originally Posted by Vanilla
I think I got it.
[LaTeX ERROR: Convert failed] can be simpified to [LaTeX ERROR: Convert failed] , which then differentiates to [LaTeX ERROR: Convert failed] . And for [LaTeX ERROR: Convert failed] , [LaTeX ERROR: Convert failed] . 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.