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# Optimization Calculus

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A rectangular storage container with an open top is to have a volume of 10 m^3. The length of this base is twice the width. Material for the base costs \$5 per square meter. Material for the sides costs \$3 per square meter. Find the cost of materials for the cheapest such container.

Apr 8, 2019

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The area of the base is   W * 2W  =  2W^2

The volume of the container = Base area * height....which implies that

10 = 2W^2 * H   ⇒    H =  10 / [ 2W^2 ]  =   5 /W^2

So....the total surface area is given by.....area of the base + side area  =

2W^2 +  2(5/W^2) [ W + 2W]   =

2W^2  + (10/W^2}[ 3W]  =

2^W^2  + 30/W

So....the cost, C, to be minimized is  this :

C =  (base cost of materials)(base area) + (side cost of materials)(side area)

C = 5(2W^2)  + 3(30/W)

C = 10W^2  + 90/W

I can do this in 2 ways

(1) Calculus.....take the derivative of the cost....set to 0  and solve

So we have

C'  = 20W  - 90/W^2  = 0

20W   = 90/W^2

W^3  = 90/20

W^3  = 9/2

W^3 = 4.5

W = (4.5)^(1/3)  ≈ 1.651

Subbing this back into the cost function, the minimized cost is

C  = 10[(4.5)^(1/3)] ^2  + 90/[4.5]^(1/3)  ≈  \$81.77

(2)  If you haven't had Calculus...we can use a graph here : https://www.desmos.com/calculator/e0v6iwu5bo

Note that the function's minimized cost is ≈  \$81.77   Apr 8, 2019