Question

Help with optimization problems!

Determine the dimensions of a rectangular solid (with a square base) with maximum volume if its surface area is 325 meters.

Answer 1

So we want to `maximize volume` given a `constraint on the surface area`.

So we'll need to come up with equations for the Volume and Surface Area.

Answer 2

I really have issues with these word / geometry problems; barely passed geo. years ago, lol. But, let me see if I can fiind the formulas. :D

Answer 3

I'll draw a picture, maybe that will help.
Because we kind of have to come up with the formulas on our own for this one.

Answer 4

Oh.That could be a problem for me. :| But, go on, I'll pay attention. I really want to get this stuff down.

Answer 5

So what we're doing with this rectangular box is .. we're labeling the sides as variables.
The idea is, we want to find out which variables (lengths) each side should be, to maximize volume (given the constraint on the surface area).

Answer 6

We have to label ~ `length` `width` and `height`.
The reason you see x TWICE is because they told us the base is SQUARE.
Meaning the `length` and `width` will be the same.

Answer 7

If that is confusing, we can use the letters that you're maybe more familiar with.

The volume of a rectangle would be,
\(\large V=lwh\)
right? Length times width times height.
In this case, the length and width are equal, so our equation is,
\(\large V=l^2h\)

Which way makes more sense? The x/y or the l/h? :O
Which method should we use?
It's just labeling so it doesn't matter to me.

Answer 8

Let's do x and y; it's not confusing for me, plus, I know that for the maximizing thing x and y are generally used. :) Sorry, If I'm quiet, I'm copying this down as we're going.

Answer 9

We need to establish another equation, one for Surface Area.
The Surface Area is the SUM of the area of all of faces of the rectangle.

The base has an area of \(\large x \times x\).
But there is another panel with this same area. The top panel.

So we'll have 3 pairs of panels (6 total sides) that we're adding together to get the Surface Area.

Answer 10

Here is the equation I'm coming up with.
Lemme know if it's still confusing where I got some of this from.

\[\large A=2x^2+2xy+2xy\]

Answer 11

This is a tricky problem. There are a lot of little pieces to take care of.
We've established our two equations we'll need.\[\large V=x^2y\]\[\large A=2x^2+4xy\]

They told us that the Surface Area is constrained to \(\large 325m\).
Let's plug that into our Surface Area function.\[\large 325=2x^2+4xy\]

Answer 12

Our goal right now is to get our Volume function in terms of ONE VARIABLE so we can take it's derivative without any problems.

Right now it's in terms of \(\large x\) and \(\large y\).
We're going to use our Constraint equation, solve for \(\large y\) and then plug that into our Volume function.

Answer 13

\[\large 325=2x^2+4xy\]So we want to solve this for \(\large y\).
Hmm what do you get? :)

Answer 14

Still here, just taking notes; give me a moment. :D

Answer 15

brb, gonna make some chocolate milk :O

Answer 16

:O

Answer 17

I may have done this wrong but, I got: \[y=-\frac{ 1 }{ 2 }x+\frac{ 325 }{ 4x }\]

Answer 18

Ok looks good. But let's not simplify it that much.
We're going to need to plug it into the Volume Function, so I think it will be easier to work with if it's in this form,\[\large y=\frac{325-2x^2}{4x}\]

Answer 19

Oh, okay.

Answer 20

\[\large \color{orangered}{y=\frac{325-2x^2}{4x}}\]

\[\large V=x^2\color{orangered}{y} \qquad \rightarrow \qquad V=x^2\color{orangered}{\large \color{orangered}{\left(\frac{325-2x^2}{4x}\right)}}\]

Answer 21

Which, if you simplify that down, I think you get something like,\[\large V=\frac{325}{4}x-\frac{1}{2}x^3\]

Answer 22

That's what I got. :)

Answer 23

So the hard part is over:
~We set up our two equations.
~We got our equation in terms of one variable.


Now we need to maximize the volume function.
We'll take the derivative of the volume function and set it equal to 0.
This will give us x-values that either maximize or minimize the volume.

Answer 24

What do you get for your x-values? :O

Answer 25

Okay, so the derivative of the volume function would be:\[V'(x) = \frac{ 325 }{ 4 }-\frac{ 3 }{ 2 }x^2\]

Answer 26

Looks good. So now we need to find critical points,\[\large 0=\frac{325}{4}-\frac{3}{2}x^2\]

Answer 27

I just popped into my calculator to make it easier, and I got:
\[x=\pm \frac{ 5\sqrt{78}}{ 6 }\]

Answer 28

Hmm calculator is giving you some weird numbers :O

We should get,\[\large x=\pm \sqrt{\frac{325}{6}} \qquad \qquad \qquad x=\pm5\sqrt{\frac{13}{6}}\]

Answer 29

Hmm. o.o let me re-do that.

Answer 30

Hm, it's till coming up with the same thing. Weird...

Answer 31

How did you factor that out, actually? :o

Answer 32

@zepdrix -- Oh, have no idea how the calculator failed, but I seee how you got that. xD