Question 61442
x = side of square base in feet
h = height of box in feet
cost of base = {{{.30x^2}}}
cost of 4 sides = {{{.10*4*h*x}}}
cost of tgop = {{{.20x^2}}}
{{{C = .3x^2 + .4xh + .2x^2}}}
{{{(x^2)h = 20}}} volume of box = 20 ft3
{{{h = 20/x^2}}}
{{{C = .3x^2 + .4x*(20/x^2) + .2x^2}}}
{{{C = .3x^2 + 8/x + .2x^2}}}
{{{C = .5x^2 + 8/x}}}
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
{{{x^2h = 20}}}
{{{4h = 20}}}
{{{h = 5}}}
2x2x5 are the dimensions for min cost = $6.00