calculus
posted by Bob .
A squarebased, boxshaped 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?

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
Cb=C1*x²
Cost of the four sides is
Cs=C2*(4x*h)
Cost of the cover is
Cc=C3*x²
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. 
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