##### determine the dimensions of the box that can be constructed at minimum cost.

label Calculus
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Jul 16th, 2015

A rectangular box is to have a square base and a volume of 20 feet cubed. If the material for the base costs \$.30/square foot, the material for the sides costs \$.10/square foot, and the material for the top costs \$.20/square foot, determine the dimensions of the box that can be constructed at minimum cost
:
Let y = the cost of the box
:
Let x = one side of the square base
:
Let h = height of the box
:
We know that the box has to be 20 cu ft:
x^2*h = 20
h = (20/x^2)
:
Costs:
Base = .30(x^2)
Top = .20(x^2)
1 side = .10(x*(20/x^2)) = .10(20/x)
4 sides = .40(20/x) = 8/x
:

Total cost:
y = Base + Top + 4 sides
y =.30(x^2) + .20(x^2) + 8/x
y = .5x^2 + 8/x
Graph it:

:
Looks like min cost (y) occurs when x = 2 ft, looks like a cost(y) of \$6.00
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Then the height = 20/2^2 = 5 ft
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What is the cost then?
top .30(2^2) = 1.20
bot .20(2^2) = 0.80
hth .10(2*5) = 1.00 * 4 sides = \$4.00
Total cost = \$6.00 like the graph indicates.

:
Check: 2 * 2 * 5 = 20 cu ft as required

x = side of square base in feet
h = height of box in feet
cost of base = [img src="http://www.algebra.com/cgi-bin/plot-formula.mpl?expression=.30x%5E2&x=0003" align="MIDDLE" alt=".30x%5E2" >
cost of 4 sides =
cost of tgop =

volume of box = 20 ft3

I make a table
x = 1
C = 8.5
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x = 2
C = 6
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x = 3
C = 7.17
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x = 4
C = 10
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It looks like x = 2 may give a minimum cost
I tried x = 1.8 and x = 2.2
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x = 1.8
C = 6.06
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x = 2.2
C = 6.06
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so, C=6 really is a min

2x2x5 are the dimensions for min cost = \$6.00

Please let me know if you nea case for torture by michale levined any clarification. I'm always happy to answer your questions.
Jul 16th, 2015

i am vert confused as to what the answers are

Jul 16th, 2015

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Jul 16th, 2015
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Jul 16th, 2015
Oct 23rd, 2017
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