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- anonymous

A box with an open top is to be constructed out of a rectangular piece of cardboard with dimensions length=8 ft and width=9 ft by cutting a square piece out of each corner and turning the sides up. Determine the length x of each side of the square that should be cut which would maximize the volume of the box.

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- anonymous

- schrodinger

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- radar

Let x equal the dimension of the corner that is going to be cut out (so the sides fold up and make a box. This x will also equal the height of the box.
The dimensions of the box will then be Length=8-2x (corners cut out at each end.)
width =9-2x. Volume will equal l*w*h. Please review this and see if you follow with understanding.

- radar

\[V=(9-2x)(8-2x) x\]

- anonymous

Following you so far. haha, that's about as far as I got on my own though. lol

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- radar

\[V=(72-34X+4X ^{2})x\]

- radar

\[V=72x-34x ^{2}+4x ^{3}\]

- radar

Rearranging in to a more standard form\[V=4x ^{3}-34x ^{2}+72x\]
Are you still with me?

- anonymous

Yep. Then we take the derivative?

- radar

Yes and set to zero and then solve for x.

- radar

Let me know what you get.

- anonymous

Derivative of.. \[12x ^{2} -68x +72\] ?

- radar

Yes, to make it easier set it to zero and divide both sides by 4 getting:

- radar

\[3x ^{2}-17x+18=0\]

- anonymous

so I'm gettinggg.. \[1/6 (17\pm \sqrt{73}\]

- anonymous

with another ) at the end.. lol

- radar

You're ahead of me. I was trying to factor before going for the quadratic equation. It looks like that was the way to go. I'll check it out and if there is an error I will be back. Remember this is the dimension of the corner cut out, to get the dimensions of the box you will subtract twice this from the 8' and 9'

- anonymous

How do I know which x value to use though, the plus gives me approximately 1.4 and the minus gives me approximately 4.3

- anonymous

Nevermind I've got it. haha thank you!

- radar

Good mic61 your computation was correct. Good day.

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