Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

Yep. Then we take the derivative?

Answer 8

Yes and set to zero and then solve for x.

Answer 9

Let me know what you get.

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

with another ) at the end.. lol

Answer 15

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'

Answer 16

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

Answer 17

Nevermind I've got it. haha thank you!

Answer 18

Good mic61 your computation was correct. Good day.