(a) What size squares should be cut from the corners to obtain a box of the maximum volume. (b) What are the sides of the square if k = 12 inches?

Question

A square sheet of cardboard of side k is used to make an open box by cutting squares of equal size from the 4 corners and folding up the sides.

(a) What size squares should be cut from the corners to obtain a box of the maximum volume.

(b) What are the sides of the square if k = 12 inches?

Answer 1

Let x be the side of the squares to be cut out.

The resulting box will have as base a square of side (k - 2x)

and will have height x.

It will have volume V = x(k - 2x)² = 4x³ - 4kx² + k²x

Physically meaningful values of x lie in the interval [0, k/2].

To find local extrema we set the derivative to 0

12x² - 8kx + k² = 0

(a) The solutions to the above quadratic equation are

x = (8k +- sqrt(64k² - 48k²))/24

which gives two solutions:

x = k/2 and x = k/6

As candidates for absolute maximum we thus have these two local extrema plus

the bounds of the interval [0, k/2].

That gives us three candidates: 0, k/6, k/2

We can plug them into the equation of volume to see which gives the

largest value, or we can see it simply from physical considerations:

Both x = 0 and x = k/2 would give us a box of volume 0,

while x = k/6 gives us non-zero volume.

Therefore x = k/6 is the answer.

(b) For k = 12, maximum volume is obtained with x = 12/6 = 2 inches.