Maximum volume that can be calculated?

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Discussion Overview

The discussion revolves around calculating the maximum volume of a box given a specific surface area constraint. Participants explore different interpretations of the surface area equation and the implications for volume calculation, with a focus on mathematical reasoning and problem-solving techniques.

Discussion Character

  • Mathematical reasoning
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant proposes a surface area equation of 7200 = 6x^2 + 4xy and calculates a maximum volume of 540,000.
  • Another participant challenges the surface area equation, suggesting it should be 7200 = 4xy + 2x^2, and finds a maximum volume of 24,000 instead.
  • A later reply provides a detailed derivation of the volume based on the corrected surface area equation, confirming the maximum volume as 24,000.
  • Participants discuss a misunderstanding regarding the problem's context, clarifying that costs associated with covering different parts of the box influence the surface area equation.
  • One participant acknowledges a mistake in their calculations and realizes their derived volume matches the others' findings.
  • Another participant points out a consistent error in terminology, where "area" was mistakenly used instead of "volume."

Areas of Agreement / Disagreement

There is no consensus on the initial surface area equation, as participants present competing interpretations. However, there is agreement on the maximum volume being 24,000 after corrections are made.

Contextual Notes

The discussion highlights the importance of accurately interpreting problem statements and equations, as well as the potential for small mistakes to lead to significant discrepancies in results.

Caldus
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If the surface area equation for a box is given as 7200 = 6x^2 + 4xy where x represents the length and width of the box while y represents the height, then what is the maximum volume that can be calculated?

My answer was 540000, but not sure.
 
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Did you come up with the equation on your own?

Because I would think that, with length and width being x and y being height, the surface area would be 7200 = 4xy + 2x^2.

Anyway, with what you gave, I found the max. volume to be something else...

Volume: V = x^2 * y
Surface Area: 7200 = 6x^2 + 4xy

Solve surface area for y: (7200 - 6x^2)/4x, simplify it a bit, (3600 - 3x^2)/2x

Substitute it into the volume...

V = x^2 * (3600 - 3x^2)/2x
V = [x(3600 - 3x^2)]/2
V = (3600x - 3x^3)/2
V = 1800x - (3/2)x^3

Take the derivative...
V' = 1800 - (9/2)x^2

Set equal to 0 and solve for x...
1800 - (9/2)x^2
1800 = (9/2)x^2
3600 = 9x^2
400 = x^2
x = +/- 20

And, because we can't have a negative length, x = 20.

Next, find y by going back into the surface area equation...

7200 = 6x^2 + 4xy
7200 = 6(20)^2 + 4(20)y
7200 = 2400 + 80y
4800 = 80y
60 = y

Now, back into the volume...
V = x^2 * y
V = (20)^2 * 60
V = 400 * 60
V = 24000

Someone may want to double check that as I always make some mistake to screw up my work.
 
I got 24,000...
 
nope, its fine, i got the same thing
 
How did you get 540,000?
 
Oh my God, I realized that I made a stupid little mistake. I will show you where I messed up:

OK, let me give the actual details of the problem. Maybe I misinterpreted the problem itself.

You have $7200. It will cost you $1 per square meter to cover the bottom part of a box. It will cost you $5 per square meter to cover the top part of a box. For every other side, it will cost you $1 per square meter to cover it. Give the maximum possible volume for this box with the amount of money you have.

So my equation ends up being: 7200 = 4xy + x^2 + 5x^2 which simplifies to 7200 = 4xy + 6x^2
Then I calculated the area to be x^2y.

In the surface area equation, I solve for y, so area ends up equaling:

x^2(((7200 - 6x^2)/4x))

I simplified to:

(7200x^2 - 6x^4)/4x
1800x - (3/2)x^3

Then I got the derivative of area:

1800 - (9/2)x^2

Set it to 0 and find the max value:

0 = 1800 - (9/2)x^2
-1800 = -(9/2)x^2 <--- Right here I accidently interpreted the coefficient to be 3 because for some reason I was thinking (6/2)x^2 instead of (9/2)x^2.
x = 20

Plug 20 back in the original area equation with only x's and you get 24,000.

And I apparently I got the same answer as you guys now that I realize my mistake.

Uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuugggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg! Why did I have to do this on my test! I know this stuff! Gaaaaaaaaah!

Sorry. ; )
 
Glad you got the right answer.

By the way, in your last response you were consistently saying "area" when you meant "volume".
 
Yep, you're right.

I have always been very good about making little mistakes like the one you just mentioned as well as the mistake I made in the problem. My test grades would be so much better otherwise, LOL.
 

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