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# Calculus/Finding minimum costs

Question
A closed crate with a square base is to have a volume of 250 feet.
Material for top and bottom costs: \$2/sq foot
Material for the sides costs: \$1/sq foot.
Find the minimal material cost

Thanks

Let s be the length of a side of the square bottom and let h be the height of the box.

Volume = (s)(s)(h) = 250

(s^2)(h) = 250

h = 250s^-2

area for the botton = s^2

area for the top = s^2

area for each of the four sides = sh = (s)(250s^-2) = 250s^-1

Total cost = 2s^2 + 2s^2 + (4)(1)(250s^-1)

= 4s^2 + 1000s^-1

Let C(s) =  4s^2 + 1000s^-1

C'(s) = 8s - 1000s^-2

0 = 8s - 1000s^-2

0 = 8s^3 - 1000

s^3 = 1000/8

s^3 = 125

s = 5

The first derivative test shows that the total cost is a minimum when s = 5

minimal material cost is

C(5) = (4)(5^2) + (1000) 5^-1

= (4)(25) + 1000/5

= 100 + 200

= 300

minimal cost is \$300

Calculus

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I can answer questions from the standard four semester Calculus sequence. I am not prepared for questions on Tensor Calculus. Everything else is welcome. Derivatives, partial derivatives, ordinary differential equations, single and multiple integrals, change of variable, vector integration (Green`s Theorem, Stokes, and Gauss) and applications.

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