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A square-based, box-shaped shipping crate is designed to have a volume of 16ft^3. The material used to make the base costs twice as much (per ft^2) as the material in the sides, and the material used to make the top costs half as much (per ft^2) as the material in the sides. What are the dimensions of the crate that minimize the cost of materials?

  • calculus -

    Since the base is square, and the volume is constant, there is only one variable, such as the side of the base, x.

    The height is therefore h=V/x².

    The cost of the base is
    Cost of the four sides is
    Cost of the cover is

    Total cost as a function of x
    C(x)= C1*x²+C2*(4x*V/x²)+C3*x²
    =(C1+C3)x²+ 4C2*V/x

    Differentiate with respect to x and equation f'(x) to zero. Solve for x0 where the cost is maximum/minimum.

    To find out if f(x0) is a maximum or minimum, calculate the second derivative and evaluate f"(x0). If f"(x0)>0, f(x0) is a minimum. If f"(x0)<0, f(x0) is a maximum.

  • calculus -


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