SOLUTION: An open box with a square base is required to have a volume of 10 cubic feet. Assume the box is to be made from a square piece of cardboard that is has original dimensions (x + 2h)

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Question 356869: An open box with a square base is required to have a volume of 10 cubic feet. Assume the box is to be made from a square piece of cardboard that is has original dimensions (x + 2h)-by-(x + 2h) by cutting out 4 h-by-h squares on the corners of the cardboard and folding up the sides where h is the height of the open box sides (see figure below)
Base length x h by h square cut out to form open box

A] Express the surface area A(x) of the box as a function of the length of the base x.
B] Use a graphing utility to graph A(x) and determine the dimensions of an open box with the smallest surface area possible.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
An open box with a square base is required to have a volume of 10 cubic feet.
Assume the box is to be made from a square piece of cardboard that is has
original dimensions (x + 2h)-by-(x + 2h)
by cutting out 4 h-by-h squares on the corners of the cardboard and folding
up the sides where h is the height of the open box sides
Base length x h by h square cut out to form open box
:
The removal of the h by h corners, reduces the dimension by 2h, therefore
The box dimensions will x by x by h, therefore:
Vol = x^2*h
Given the vol as 10 cu/ft:
x^2*h = 10 cu/ft
h = 10%2Fx%5E2
:
A] Express the surface area A(x) of the box as a function of the length of the base x.
Surface area of a 5 sided (open) box:
S.A. = x^2 + 4(x*h)
replace h with 10%2Fx%5E2
S.A. = x^2 + 4(x*10%2Fx%5E2)
Cancel x
S.A. = x^2 + 4(10%2Fx)
S.A = x^2 + 40%2Fx; the surface area as a function of x
:
B] Use a graphing utility to graph A(x) and determine the dimensions of an open box with the smallest surface area possible.
In a TI83 or similar enter y= x^2+40%2Fx, I used a scale: -2,+10; -10,+50
Looks like this:
+graph%28+300%2C+200%2C+-2%2C+10%2C+-10%2C+50%2C+x%5E2%2B%2840%2Fx%29%29+
Using the min feature on the calc, I got x=2.744, surface area of 22.1 is min
:
Using x=2.744,
h = 10%2F2.744%5E2
h ~ 1.33 is the height
Find the Surface area from these values
S.A. = 2.744^2 + 4(2.744*1.334)
S.A. = 7.53 + 4(3.65)
S.A. = 7.53 + 14.6
S.A. = 22.13 which agrees with the Calc values
:
Did all this make sense to you?