MHB How to Maximize the Volume of a Cylinder with a Given Perimeter?

[Noah1]

it is given that the perimeter of the rectangle is (80 + 120 + 80 + 120) = 400 cm From this you need to make a cylinder with maximin volume:
400 = 2r + 2h
2h = 400 - 2r
h = 200 - r .
We wish to MAXIMIZE the total VOLUME of the resulting CYLINDER
V = πr^2 h
However, before we differentiate the right-hand side, we will write it as a function of r only. Substitute for h getting
V = πr^2 h
V = πr^2 (200-r)
V = π(200r^2-r^3 )
Now differentiate this equation, getting
V’ = π(200r^2-r^3 )
V’ = π(400r-3r^2 )

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it is given that the perimeter of the rectangle is (80 + 120 + 80 + 120) = 400 cm From this you need to make a cylinder with maximin volume:
400 = 2r + 2h
2h = 400 - 2r
h = 200 - r .
We wish to MAXIMIZE the total VOLUME of the resulting CYLINDER
V = πr^2 h
However, before we differentiate the right-hand side, we will write it as a function of r only. Substitute for h getting
V = πr^2 h
V = πr^2 (200-r)
V = π(200r^2-r^3 )
Now differentiate this equation, getting
V’ = π(200r^2-r^3 )
V’ = π(400r-3r^2 )

At this point I don't know where to go from here to solve for r

 

[MarkFL]

If the dimensions of the rectangle are already given, then we only have 2 choices:

$$V_1=\pi\left(\frac{80}{2\pi}\right)^2120$$

$$V_2=\pi\left(\frac{120}{2\pi}\right)^280$$

 

[soroban]

It is given that the perimeter of the rectangle is 400 cm.
From this you need to make a cylinder with maximin volume:

I interpret this problem as follows:

There is a sheet of paper with a perimeter of 400 cm.
We will "roll" the sheet into a cylinder.
Find the dimensions of the paper which produces the cylinder of maximum volume.

This version requires a bit more work . . .

 

[MarkFL]

If we are free to choose the dimensions of the rectangle, so long as the perimeter is a certain value, then consider the diagram:

http://www.ekshiksha.org.in/Image_Surface_Areas_and_Volumes_IX/8.png

Let $S$ be the semi-perimeter (a constant), so that we have:

$$l+h=S$$

The volume of the resulting cylinder is:

$$V=\pi\left(\frac{l}{2\pi}\right)^2h=\frac{1}{4\pi}l^2h$$

At this point we may choose which of the variables we wish to replace, and since $l$ is being squared, for simplicity let's replace $h$:

$$V=\frac{1}{4\pi}l^2(S-l)=\frac{1}{4\pi}\left(Sl^2-l^3\right)$$

Hence, we now have the volume as a function of one variable $l$:

$$V(l)=\frac{1}{4\pi}\left(Sl^2-l^3\right)$$

We may ignore the constant $$\frac{1}{4\pi}$$ and simply optimise:

$$f(l)=Sl^2-l^3$$

So, taking the first derivative, and equating to zero, we obtain:

$$f'(l)=2Sl-3l^2=l\left(2S-3l\right)=0$$

From this, we obtain the two critical values:

$$l=0,\,\frac{2S}{3}$$

Now, if we utilize the second derivative test, on computing $f''$, we obtain:

$$f''(l)=2S-6l$$

And we find:

$$f''(0)=2S>0$$ This critical value is at a relative minimum.

$$f''\left(\frac{2S}{3}\right)=-2S<0$$ This critical value is at a relative maximum.

So, in order to maximize the volume of the cylinder, we require:

$$l=\frac{2S}{3}$$

$$h=\frac{S}{3}$$

Does that make sense?