SOLUTION: a company wishes to manufacture a box with a volume of 36 cubic feet that is open on top and is twice as long as it is wide. find the dimensions of the box produced from the minim

Algebra ->  Volume -> SOLUTION: a company wishes to manufacture a box with a volume of 36 cubic feet that is open on top and is twice as long as it is wide. find the dimensions of the box produced from the minim      Log On


   

Question 443207: a company wishes to manufacture a box with a volume of 36 cubic feet that is open on top and is twice as long as it is wide. find the dimensions of the box produced from the minimum amount of material.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
On thinking this over last night, I discovered a big error in this, let me correct it here
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a company wishes to manufacture a box with a volume of 36 cubic feet that is open on top and is twice as long as it is wide.
find the dimensions of the box produced from the minimum amount of material.
Let x = the width
Let 2x = the length
Let h = the height
then
vol = x*2x*h
so we have
2x^2*h = 36
h = 36%2F%282x%5E2%29
h = 18%2Fx%5E2
Surface area: two ends + 1 bottom + 2 sides (no top)
S.A. = 2(x*h) + 1(2x*x) + 2(2x*h)
:
I am truly sorry for this error. Carl
:
S.A. = 2xh + 2x^2 + 4xh
S.A. = 2x^2 + 6xh
Replace h with 18%2Fx%5E2
S.A = 2x^2 + 6x(18%2Fx%5E2)
S.A = 2x^2 + 6(18%2Fx)
S.A = 2x^2 + (108%2Fx)
Graph this equation to find the value of x for minimum material
+graph%28+300%2C+200%2C+-1%2C+5%2C+-20%2C+150%2C+2x%5E2%2B%28108%2Fx%29%29+
Min surface area when x = 3.0 is the width
then
2(3) = 6 is the length
;
Find the height:
h = 18%2F3%5E2
h = 2
:
Box dimensions for min surface area: 3 by 6 by 2; much better numbers
:
Check the vol of these dimensions: 3*6*2 ~ 36