Precalc question

Question

Steve likes to entertain friends at parties with “wire tricks.” Suppose he takes a piece of wire 60 inches long and cuts it into two pieces. Steve takes the first piece of wire and bends it into the shape of a perfect circle. He then proceeds to bend the second piece of wire into the shape of a perfect square. Where should Steve cut the wire so that the total area of the circle and square combined is as small as possible? What is this minimal area? What should Steve do if he wants the combined area to be as large as possible?

Answer 1

Assign variables to the 2 pieces: X and Y. Use X for the square and Y for the circle

Write equations using these variables and the information provided.

X + Y = 60

The square: each side is X/4. The area is (X/4)² = X² / 16

The circle: Y is the circumference, so the radius is Y/(2*pi). The area is pi*r² = pi*(Y² /4*pi² ) = Y² /4*pi

We can now write an equation for the total area as a function of either X or Y. Let's use X----this means use 60 - X instead of Y.......So: the total area is

A = X² / 16 + 1/4pi (60 - X)²

Expand and Simplify:

A = X² / 16 + 1/4pi (3600 - 120X + X² )

A = (1/16 + 1/4pi)X² - (120/4pi)X + 900/pi

A = 0.142X² - 9.55X + 286

Recognize that this is a parabola that "opens up". You can convert this to the standard form to find the vertex, or simply take the 1st derivative.

Using the 1st derivative:

A' = 0.282X - 9.55

To find the minimum, set A' = 0, and solve to get X = 33.9

The maximum will occur at either X = 0 or X = 60

If X = 0, A = 3600/4pi = 286

If X = 60, A = 3600/16 = 225

So, the maximum area is when all the wire is used in a circle