A 6
′′
× 6
′′
square sheet of metal is made
into an open box by cutting out a square at
each corner and then folding up the four sides.
Determine the maximum volume, Vmax, of the
box.
1. Vmax = 31 cu. ins.
2. Vmax = 16 cu. ins.
3. Vmax = 21 cu. ins.
4. Vmax = 36 cu. ins.
5. Vmax = 26 cu. ins.

Question

A 6
′′
× 6
′′
square sheet of metal is made
into an open box by cutting out a square at
each corner and then folding up the four sides.
Determine the maximum volume, Vmax, of the
box.
1. Vmax = 31 cu. ins.
2. Vmax = 16 cu. ins.
3. Vmax = 21 cu. ins.
4. Vmax = 36 cu. ins.
5. Vmax = 26 cu. ins.

.sam. · Answer

Finding the derivative of a function, say f(x), leads to finding the slope of the tangent line at x. If the slope is 0, then you have the possibility of a minimum or a maximum at x.

So, you need to find an equation for the volume of the box, take the
derivative, set it equalt to zero, solve for x, then use the second derivative (or a simple graph) to tell if the value you get is a minimum or maximum, or simply an inflection point.

If you draw a square with a notch of sixe x removed fom each corner, you will see that what is left of each side is just 6 - 2x. Stare at the picture for a moment, and it will become obvious to you that the volume of the box is the area of the base, which is (6 - 2x) on a side, times the height of the walls, which is just x.

So, the volume of the box is simply

volume = height times area of base

v = x(6 - 2x)^2

v = 36x - 24x^2 + 4x^3

Find v' then solve for v' = 0, draw a graph of v' to convince your self that at least one of your solutions is a max, and you are done.