SOLUTION: I have a piece of cardboard, twice as long as it is wide. I will cut 4x4 suqares out of each corners, so I fold up the corners and use it as a tray. The volume of the tray will be

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Question 633011: I have a piece of cardboard, twice as long as it is wide. I will cut 4x4 suqares out of each corners, so I fold up the corners and use it as a tray. The volume of the tray will be 96 cubic inches. How large is the piece of cardboard?
Answer by graphmatics(170) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the width of the cardboard. 2x is the length of the card. After cutting out the 4 by 4s on the corners and creating the tray the area of the bottom of the tray will be (x-8)(2x-8). The height of the tray is 4. The volume of the tray is (x-8)(2x-8)4. Because the volume of the tray is 96 we have that (x-8)(2x-8)(4)=96 so %282%2Ax%5E2-24x%2B64%29%284%29=96. 2%2Ax%5E2-24x%2B64=24. x%5E2-12x%2B32=12. x%5E2-12x%2B20=0. We solve this quadratic equation.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-12x%2B20+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-12%29%5E2-4%2A1%2A20=64.

Discriminant d=64 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--12%2B-sqrt%28+64+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-12%29%2Bsqrt%28+64+%29%29%2F2%5C1+=+10
x%5B2%5D+=+%28-%28-12%29-sqrt%28+64+%29%29%2F2%5C1+=+2

Quadratic expression 1x%5E2%2B-12x%2B20 can be factored:
1x%5E2%2B-12x%2B20+=+1%28x-10%29%2A%28x-2%29
Again, the answer is: 10, 2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-12%2Ax%2B20+%29
. We must select x=10 because x must be greater than 8. The cardboard dimensions are 10 and 20.