MinMax01 Design an open-top box.

Design an open-top box for maximum volume. The box is to be made from a square piece of material of dimension a. What size square should be cut from each corner to make the box?

The side of the square taken from each corner is x. After the corner pieces are removed, the box is formed by bending the sides along the lines indicated.

The defining equation is V = (a - 2x)² x.

The bottom of the box is a - 2x by a - 2x and the height is x. The a is a constant, making the equation for V one with only one variable, x. Multiplying, we have the following:

V = (a² - 4ax + 4x²)x = 4x³ - 4ax² + a²x

Differentiate V, and set the derivative equal to zero to find the maxima and minima of the curve of V vs. x.

V' = 12x² - 8ax + a² = a² - 8ax + 12x² = (a - 6x)(a - 2x)

Setting V' = 0 produces values for x of a/6 and a/2. These are the maxima or minima. The value a/2 is obviously the minimum since this is a box of zero volume! The value a/6 must be the maximum. The second derivative test will tell for sure. The second derivative of V is V'' = 24x - 8a. At x = a/2, V'' = 12a - 8a = 4a (positive or minima), and at x = a/6, V'' = 4a - 8a = -4a (negative or maxima).

Maximum volume occurs when the square piece removed from the edge of the original square is one-sixth the length of the side.