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10 October, 05:32

This is the same bonus problem as Lesson 12. This time, you have to use the method of Lagrange Multipliers to solve it. A rectangular tank with a bottom and sides but no top is to have volume 500 cubic feet. Determine the dimensions (length, width, height) with the smallest possible surface area.

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  1. 10 October, 05:37
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    length=10 ft, width = 10 ft and height = 5 ft

    Step-by-step explanation:

    using Lagrange multipliers, we have the main function that is the Area A of the tank,

    A (x, y, z) = x*y + 2*x*z + 2*y*z

    constrained to the Volume V (x, y, z) = x*y*z=a = 500 ft³

    using Lagrange multipliers

    Ax - λ*Vx = 0 → (y + 2*z) - y*z*λ = 0 → λ = 1/z + 2/y

    Ay - λ*Vy = 0 → (x + 2*z) - x*z*λ = 0 → λ = 1/z + 2/x

    Az - λ*Vz = 0 → (2*x + 2*y) - x*y*λ = 0 → λ = 2/x + 2/y

    V = a → x*y*z=a

    adding the first and second equations

    2*λ = 1/z + 2/y + 1/z + 2/x = 2/z + λ

    λ = 2/z → z = 2/λ

    therefore

    λ = 1/z + 2/y = λ/2 + 2/y

    λ/2 = 2/y → y = 4/λ

    and similarly x=4/λ

    then

    x*y*z=a

    2/λ*4/λ*4/λ = a

    32/λ³ = a

    λ = ∛32/a

    therefore

    z = 2/λ = 2*∛a/32 = 2*∛ (500/32) = 5

    y = 4/λ = 2*5 = 10

    x=10

    therefore the dimensions that minimize the area are x=10, y=10 and z=5 (length=10 ft, width = 10 ft and height = 5 ft)
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