SOLUTION: Hi,
I am having trouble solving the problem below and could use some assistance:
An open box is to be made from a rectangular metal sheet by cutting two inch squares from eac
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-> SOLUTION: Hi,
I am having trouble solving the problem below and could use some assistance:
An open box is to be made from a rectangular metal sheet by cutting two inch squares from eac
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Question 815495: Hi,
I am having trouble solving the problem below and could use some assistance:
An open box is to be made from a rectangular metal sheet by cutting two inch squares from each of the corners and folding up the sides. The perimeter of the sheet of metal is 100 inches. Which dimensions of the box will provide the maximum area for the base of the box?
Thank you so much for your help! Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! An open box is to be made from a rectangular metal sheet by cutting two inch squares from each of the corners and folding up the sides.
The perimeter of the sheet of metal is 100 inches.
Which dimensions of the box will provide the maximum area for the base of the box?
The dimensions of the sheet of metal:
2L + 2W = 100
Simplify, divide by 2
L + W = 50
then
L = (50-W)
;
The length & width of the box will be 4" less than the Length & Width of the Sheet
(L-4) by (W-4)
We know the height of the box is 2 inches
:
The volume of the open box
V = (L-4)*(W-4)*2
Replace L with (50-W)
V = ((50-W)-4) * (W-4) * 2
V = (-W+46) * (W-4) * 2
FOIL
V = (-W^2 + 4W + 46W - 184) * 2
V = -2W^2 + 100w - 368
Max volumee occurs at the axis of symmetry x = -b/(2a)
W =
W = 25 inch width gives max volume
then
L = 50 - 25 = 25 inch length, a square piece of sheet metal
:
The box dimensions then 21 by 21 by 2 = 882 cu/inches max volume
