Optimization Problem - # 2

Jason76

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If 10,800 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

(first use surface area equation \(\displaystyle x^{2} + 4xy = \)Surface area, then use Volume equation \(\displaystyle l * w * h\) or \(\displaystyle x^{2}y\) and then back to surface area equation after doing the derivative on the volume equation)

\(\displaystyle x^{2} + 4xy = \)Surface area

\(\displaystyle x^{2} + 4xy = 10,800\)

\(\displaystyle \dfrac{x^{2}}{4x} + \dfrac{4xy}{4x} = \dfrac{10,800}{4x}\)

\(\displaystyle \dfrac{x^{2}}{4x} + y = \dfrac{10,800}{4x}\)

\(\displaystyle y = \dfrac{10,800}{4x} - \dfrac{x^{2}}{4x} \) :confused: On the right track here?
 
Yes. Is there a reason to keep writing "\(\displaystyle \dfrac{x^2}{4x}\)" rather than just \(\displaystyle \dfrac{x}{4}\)?
 
If 10,800 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

(first use surface area equation \(\displaystyle x^{2} + 4xy = \)Surface area, then use Volume equation \(\displaystyle l * w * h\) or \(\displaystyle x^{2}y\) and then back to surface area equation after doing the derivative on the volume equation)

\(\displaystyle x^{2} + 4xy = \)Surface area

\(\displaystyle x^{2} + 4xy = 10,800\)

\(\displaystyle \dfrac{x^{2}}{4x} + \dfrac{4xy}{4x} = \dfrac{10,800}{4x}\)

\(\displaystyle \dfrac{x^{2}}{4x} + y = \dfrac{10,800}{4x}\)

\(\displaystyle y = \dfrac{10,800}{4x} - \dfrac{x^{2}}{4x} \)

\(\displaystyle y = \dfrac{10,800 - x^{2}}{4x}\) - continuing on...
\(\displaystyle x^{2}(\dfrac{10,800 - x^{2}}{4x})\)

\(\displaystyle V = x^{2}y\)

\(\displaystyle V = \dfrac{10,800x^{2} - x^{4}}{4x}\)
 
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If 10,800 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

(first use surface area equation \(\displaystyle x^{2} + 4xy = \)Surface area, then use Volume equation \(\displaystyle l * w * h\) or \(\displaystyle x^{2}y\) and then back to surface area equation after doing the derivative on the volume equation)

\(\displaystyle x^{2} + 4xy = \)Surface area

\(\displaystyle x^{2} + 4xy = 10,800\)

\(\displaystyle \dfrac{x^{2}}{4x} + \dfrac{4xy}{4x} = \dfrac{10,800}{4x}\)

\(\displaystyle \dfrac{x^{2}}{4x} + y = \dfrac{10,800}{4x}\)

\(\displaystyle y = \dfrac{10,800}{4x} - \dfrac{x^{2}}{4x} \)

\(\displaystyle y= \dfrac{10,800 - x^{2}}{4x}\)- continuing on...

\(\displaystyle x^{2}(\dfrac{10,800 - x^{2}}{4x})\) ← Where did that come from?? why did it come?

You should have written (instead of depending on our mind-reading capabilities):

\(\displaystyle V(olume) \ = \ x^{2} \ * \ \dfrac{10,800 - x^{2}}{4x}\)

Now what........
 
Last edited by a moderator:
If 10,800 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

(first use surface area equation \(\displaystyle x^{2} + 4xy = \)Surface area, then use Volume equation \(\displaystyle l * w * h\) or \(\displaystyle x^{2}y\) and then back to surface area equation after doing the derivative on the volume equation)

\(\displaystyle x^{2} + 4xy = \)Surface area

\(\displaystyle x^{2} + 4xy = 10,800\)

\(\displaystyle \dfrac{x^{2}}{4x} + \dfrac{4xy}{4x} = \dfrac{10,800}{4x}\)

\(\displaystyle \dfrac{x^{2}}{4x} + y = \dfrac{10,800}{4x}\)

\(\displaystyle y = \dfrac{10,800}{4x} - \dfrac{x^{2}}{4x} \)

\(\displaystyle y = \dfrac{10,800 - x^{2}}{4x}\) - continuing on...

\(\displaystyle V = x^{2}y\) - plug y into volume equation

\(\displaystyle x^{2}(\dfrac{10,800 - x^{2}}{4x})\)

\(\displaystyle V = \dfrac{10,800x^{2} - x^{4}}{4x}\)

\(\displaystyle V' = \dfrac{4x(21600x - 4x^{3}) - (10,800x^{2} - x^{4})(4)}{(4x)^{2}}\)

\(\displaystyle V' = \dfrac{864,000 x^{2} - 16x^{4} - 43200x^{2} + 4x^{4}}{4x^{2}}\)

\(\displaystyle V' = \dfrac{820,800 x^{2} + 20x^{4}}{4x^{2}}\)

\(\displaystyle V' = \dfrac{820,800 x^{2} + 20x^{4}}{4x^{2}} = 0\)

\(\displaystyle V' = 820,800 x^{2} + 20x^{4} = 0\)

\(\displaystyle V' = 820,800 x^{2} = -20x^{4}\)

\(\displaystyle V' = 41,000 x^{-2}\)

On the right track here?



 
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If 10,800 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

(first use surface area equation \(\displaystyle x^{2} + 4xy = \)Surface area, then use Volume equation \(\displaystyle l * w * h\) or \(\displaystyle x^{2}y\) and then back to surface area equation after doing the derivative on the volume equation)

\(\displaystyle x^{2} + 4xy = \)Surface area

\(\displaystyle x^{2} + 4xy = 10,800\)

\(\displaystyle \dfrac{x^{2}}{4x} + \dfrac{4xy}{4x} = \dfrac{10,800}{4x}\)

\(\displaystyle \dfrac{x^{2}}{4x} + y = \dfrac{10,800}{4x}\)

\(\displaystyle y = \dfrac{10,800}{4x} - \dfrac{x^{2}}{4x} \)

\(\displaystyle y = \dfrac{10,800 - x^{2}}{4x}\) - continuing on...

\(\displaystyle V = x^{2}y\) - plug y into volume equation

\(\displaystyle x^{2}(\dfrac{10,800 - x^{2}}{4x})\)

\(\displaystyle V = \dfrac{10,800x^{2} - x^{4}}{4x}\)

\(\displaystyle V' = \dfrac{4x(21600x - 4x^{3}) - (10,800x^{2} - x^{4})(4)}{(4x)^{2}}\)

\(\displaystyle V' = \dfrac{864,000 x^{2} - 16x^{4} - 43200x^{2} + 4x^{4}}{4x^{2}}\)

\(\displaystyle V' = \dfrac{820,800 x^{2} + 20x^{4}}{4x^{2}}\)

\(\displaystyle V' = \dfrac{820,800 x^{2} + 20x^{4}}{4x^{2}} = 0\)

\(\displaystyle V' = 820,800 x^{2} + 20x^{4} = 0\)

\(\displaystyle V' = 820,800 x^{2} = -20x^{4}\)

\(\displaystyle V' = 41,000 x^{-2}\)

On the right track here?.......................NO!


\(\displaystyle V' = \dfrac{820,800 x^{2} + 20x^{4}}{4x^{2}} = 0\)

\(\displaystyle 820,800 + 20x^{2} = 0\) ...............and continue....
 
Much more direct way....

\(\displaystyle V(olume) \ = \ x^{2} \ * \ \dfrac{10,800 - x^{2}}{4x} = 2700*x - \dfrac{x^3}{4}\)

\(\displaystyle V' = 2700 - \dfrac{3*x^2}{4} = 0\)

\(\displaystyle 2700 = \dfrac{3*x^2}{4}\)

\(\displaystyle x = 60\)


 
Much more direct way....

\(\displaystyle V(olume) \ = \ x^{2} \ * \ \dfrac{10,800 - x^{2}}{4x} = 2700*x - \dfrac{x^3}{4}\)

\(\displaystyle V' = 2700 - \dfrac{3*x^2}{4} = 0\)

\(\displaystyle 2700 = \dfrac{3*x^2}{4}\)

\(\displaystyle x = 60\)

Hello, bit late but this gives a different answer to your previous comment. Which one is the correct way?

Also if you go down from this: "820,800+20x2=0820,800+20x2=0 ...............and continue.... " It does not give you 60... so not sure what to do here?
 
Hello, bit late but this gives a different answer to your previous comment. Which one is the correct way?

Also if you go down from this: "820,800+20x2=0820,800+20x2=0 ...............and continue.... " It does not give you 60... so not sure what to do here?
That was a continuation of Jason's incorrect work. Discard that!

Follow the development from response 7.
 
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