Posted by MEG on Monday, November 15, 2010 at 10:03pm.
______
/_____/| $5 bottom
|| || $1 Top
|| || H $3 Side
||____||
|/____|/S
S
A = S^2 * H Cost will be related
4 = S^2 * H with Surface Area
H = 4 / S^2 So our equation is...
SA = 2S^2 + 4SH
Cost = 5S^2 + S^2 + 4*3*S(4/S^2)
Cost = 6S^2 + 48/S
Now we differentiate Cost
Cost' = 12S - 48/S^2
Now we must determine where the derivative of cost is equal to 0
0 = 12S - 48/S^2
48/S^2 = 12S
48 = 12S^3
S = 4^(1/3)
So a critical number will occur at the cube root of 4. Do a number line analysis.
Domain: [1,1.5]
-- -- ++
|----------|-----------|-------
1 1.5 4^(1/3)
The local minimum occurs at 4^(1/3), which is not in the domain situation. Instead, use 1.5 for the value of S.
H = 4/S^2
H = 4/[3/2]^2
so H = 16/9
Answer: A 1.5 x 1.5 x 16/9 box will minimize the cost.
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