SOLUTION: Im not exactly sure how to solve this: A rectangular sheet of aluminum cardboard is 5 inches longer than it is wide. if squares of 1 inch are cut from each corner and the sides

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Question 388836: Im not exactly sure how to solve this:
A rectangular sheet of aluminum cardboard is 5 inches longer than it is wide. if squares of 1 inch are cut from each corner and the sides are folded up to form a rectangular tray,
a)write an equation for the volume of the tray
b) if the volume is 30 cubic inches, find the dimensions of the cardboard to the nearest tenth.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
You end up subtracting 2 in. from each side.
Call the length of the sheet x
The width of the sheet is then x+-+5
After cutting out the 1 inch squares from the
corners, the length is x+-+2 and the
width is x+-+7
V+=+1%2A%28x+-+2%29%2A%28x+-+7%29
V+=+x%5E2+-+9x+%2B+14
And, if V+=+30 in.
30+=+x%5E2+-+9x+%2B+14
x%5E2+-+9x+-+16+=+0
Using the quadratic equation:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a+=+1
b+=+-9
c+=+-16
x+=+%28-%28-9%29+%2B-+sqrt%28+%28-9%29%5E2-4%2A1%2A%28-16%29+%29%29%2F%282%2A1%29+
x+=+%289+%2B-+sqrt%28+81+%2B+64+%29%29%2F2+
x+=+%289+%2B-+sqrt%28+145+%29%29%2F2+
x+=+%289+%2B+12.042%29%2F2
x+=+21.042%2F2
x+=+10.52
x+-+5+=+5.52
x+-+2+=+8.52
x+-+7+=+3.52
The dimensions of the cardboard are 5.52 x 10.52 in2
check:
The volume would be 1%2A8.52%2A3.52+=+29.99
This looks close enough