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Question 174983This question is from textbook california algebra 2
: 56. Tonya wants to make a metal tray by cutting four identical square corner pieces from a rectangular metal sheet. Then she will bend the sides up to make an open tray.
a. Let the length of each side of the removed squares be x in. Express the volume of the box as a polynomial function of x.
b. Find the dimensions of a tray that would have a 384-in. to the 3rd power capacity.
rectangle diagram with length 20in. and width 16in.
This question is from textbook california algebra 2
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 56. Tonya wants to make a metal tray by cutting four identical square corner pieces from a rectangular metal sheet. Then she will bend the sides up to make an open tray.
a. Let the length of each side of the removed squares be x in. Express the volume of the box as a polynomial function of x.
L = length
W = width
Vol = (L-2x)*(W-2x)*x
V = (WL - 2x*(W+L) + 4x^2)*x
V = 4x^3 - 2x^2(W+L) + WLx
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b. Find the dimensions of a tray that would have a 384-in. to the 3rd power capacity.
rectangle diagram with length 20in. and width 16in.
Do you mean to start with a sheet 20 by 16?
384 = 4x^3 - 2x^2*(16+20) +320x
384 = 4x^3 - 72x^2 +320x
x^3 - 18x^2 + 80x - 96 = 0
x = 2, 4 or 12 inches
12 won't work, since cutting 12" from 2 corners totals 24", more than the size of the sheet.
So either 2 or 4 inches will produce a tray of 384 cubic inches.
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