Question

Quick optimization question! I just need the answer for check.
If 2000 cm^2 of material is available to make a box with a square base and an open top, find the maximum volume of the box in cubic centimeters.

Answer 1

Hey zepdrix, I've rephrased the question, it does have a square base.

Answer 2

Cool :U

Answer 3

Before you do anything, can I show you what I have done so far?

Answer 4

sure ya :d

Answer 5

Wow, that is a very good box by the way.

Answer 6

lol :D

Answer 7

So anyway, I set up 2 equations.\[\large 2000 = x^2 + 4xy\]\[\large\ v = x^2y\]
I solved for y in the first equation to get
\[\frac{ 2000-x^2 }{ 4x } = y\]
Than I substituted y to get\[\large v = x^2(\frac{ 2000-x^2 }{ 4x })\]

Answer 8

Now I'm ready to take the derivative correct?

Answer 9

Yes, perfect so far!
Let's simplify a tiny bit before differentiating,\[\large\rm v=\frac14x(2000-x^2)\]

Answer 10

Ok, so I got\[\large v'=500-\frac{ 3x^2 }{ 4 }\]

Answer 11

Mmm ya looks good.

Answer 12

hmmm if i set v' to 0 and solve for x, I would get the x value of the box?

Answer 13

You would get critical values for x,
values which maximize or minimize this volume function.

If you only end up with one x as a solution,
then need to worry about what type of critical point it is.
It's probably the maximum that we're looking for.

Answer 14

then no* need to worry

Answer 15

Ohh ok, so I used a calculator and got x=25.82

Answer 16

If you're going to approximate, then yes, that looks correct.
As an exact answer it would be \(\large\rm x=\sqrt{\frac{2000}{3}}\)

Answer 17

That's the x which maximizes the volume.
What about the y? Hmm

Answer 18

Ohh, I can plug it in to the equation 2000=x^2+4xy
I got y=12.91

Answer 19

Yes, if you apply some algebra, then exact form is \(\large\rm y=\frac{\sqrt{6000}}{6}\)
But it takes a little work to come up with that one :) lol

Answer 20

So to find the maximum volume, I would do \[\large v=(\sqrt{\frac{ 2000 }{ 3 }})^2(\frac{ \sqrt{6000} }{ 6 })\approx8606.63 cm^3\]

Answer 21

Did I do it right??

Answer 22

Yayyy good job!

Here is a graph of the volume function just so you can verify:
https://www.desmos.com/calculator/4lvfaqrhd6

You can click the high point of the graph, it should be (25.82, 8606.63)

Answer 23

Oh to be fair... we actually didn't need the y value, did we?
We did all that work putting our volume formula in terms of x only,
so we could have just plugged x into that,\[\large\rm v=\frac14x(2000-x^2)\]

Answer 24

Ohh I see, we didn't need to find the y value after all. Well anyway, thanks for your help! Really appreciate it :D

Answer 25

np