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Calculus I Quick Optimization Problem

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Could you please explain this problem step by step, thank you!

You are planning to make an open rectangular box that will hold a volume of 50 cubed feet. What are the dimensions of the box with minimum surface area?

  • Calculus I Quick Optimization Problem - ,

    For symmetry reasons, the optimum must have a square base. Let the height of the walls be x and the side length be y

    Volume = x^2*y = 50
    Area = x^2 +4xy = x^2 + 4x*50/x^2
    = x^2 + 200/x
    dA/dx = 0 = 2x -200/x^2
    x^3 = 100
    x = 4.64 ft
    y = 2.32 ft

  • Calculus I Quick Optimization Problem - ,

    x = length
    y = width
    h = height
    x y h = 50
    area bottom = x y
    area sides = 2 x h + 2 y h
    total area a = x y + 2 x h + 2 y h
    so
    h = 50/xy

    a = x y + 100/y + 100/x
    da/dx=x dy/dx+y-100 dy/dx /y^2-100/x^2
    for min = 0
    dy/dx(x-100/y^2) + y-100/x^2 = 0
    x y^2 = 100 and y x^2 = 100
    y (100^2/y^4) = 100
    y^3 = 100
    y = 4.64
    x = 100/y^2 = 4.64
    h = 2.32

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