SOLUTION: A cardboard box has a square base with each edge of the box having length of x inches. The total length of all 12 edges of the box is 144in. Express the volume of the box as a func
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-> SOLUTION: A cardboard box has a square base with each edge of the box having length of x inches. The total length of all 12 edges of the box is 144in. Express the volume of the box as a func
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Question 371582: A cardboard box has a square base with each edge of the box having length of x inches. The total length of all 12 edges of the box is 144in. Express the volume of the box as a function of x. Express the height in terms of x. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Added to this solution as requested by student. Please let me know if you got
this. I am not sure when we edit a solution, that the student gets it.
ankor@att.net is my address now.
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A cardboard box has a square base with each edge of the box having length of x inches.
The total length of all 12 edges of the box is 144in.
Express the volume of the box as a function of x.
:
Given x = length of the square sides of the base
:
let h = the height of the box
:
then the 12 side total:
8x + 4h = 144
simplify, divide by 4
2x + h = 36
h = (36-2x)
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The volume:
V = x^2*h
Replace h with (36-2x)
V = x^2(36-2x)
V = -2x^3 + 36x^2, (volume in terms of x)
:
Express the height in terms of x.
Above: h = (36-2x)
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Response to student comment for length of x for max volume.
the easy way is to put the vol equation into a graph calc: y = -2x^3 + 36x^2
:
Looks something like this:
You can see max area will occur when x=12 inches, Vol will be 1728 cu/inches
:
You can see that max volume occurs when the box is a cube, 12*12*12 = 1728
(12 edges of 12 inches each)
