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October 2, 2014

October 2, 2014

Posted by **MEG** on Monday, November 15, 2010 at 10:03pm.

- Calculus -
**Cody**, Wednesday, October 19, 2011 at 10:33pm______

/_____/| $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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