A wire four feet long is cut into two pieces. One piece forms a circle of radius r, the other forms a square of side x. Choose r to minimize the sum of their areas. Then choose r t... A wire four feet long is cut into two pieces. One piece forms a circle of radius r, the other forms a square of side x. Choose r to minimize the sum of their areas. Then choose r to maximize. Read more

Steps to Solve

1. Define Variables and Constraints

Let the length of the wire be 4 feet. For a circle with radius $r$, the circumference is given by $C = 2\pi r$. For a square with side length $x$, the perimeter is $P = 4x$. We can establish the constraint:

$$ 2\pi r + 4x = 4 $$

2. Express Area in Terms of r

The area of the circle, $A_c$, is:

$$ A_c = \pi r^2 $$

The area of the square, $A_s$, is:

$$ A_s = x^2 $$

The total area, $A$, is:

$$ A = A_c + A_s = \pi r^2 + x^2 $$

3. Substitute x in Terms of r

From the constraint $2\pi r + 4x = 4$, solve for $x$:

$$ 4x = 4 - 2\pi r \implies x = 1 - \frac{\pi}{2} r $$

Now substitute this expression into the area equation:

$$ A = \pi r^2 + \left(1 - \frac{\pi}{2} r\right)^2 $$

4. Simplify the Area Expression

Expand and combine terms:

$$ A = \pi r^2 + \left(1 - \frac{\pi}{2} r\right)^2 $$

This leads to:

$$ A = \pi r^2 + 1 - \pi r + \frac{\pi^2}{4} r^2 $$

Combining like terms:

$$ A = \left(\pi + \frac{\pi^2}{4}\right) r^2 - \pi r + 1 $$

5. Find Critical Points

To minimize or maximize, take the derivative of $A$ with respect to $r$ and set it to zero:

$$ \frac{dA}{dr} = 2\left(\pi + \frac{\pi^2}{4}\right)r - \pi $$

Set the derivative equal to zero:

$$ 2\left(\pi + \frac{\pi^2}{4}\right)r - \pi = 0 $$

Solve for $r$:

$$ r = \frac{\pi}{2\left(\pi + \frac{\pi^2}{4}\right)} $$

6. Second Derivative Test

To determine if this critical point is a minimum or maximum, compute the second derivative:

$$ \frac{d^2A}{dr^2} = 2\left(\pi + \frac{\pi^2}{4}\right) $$

Since this expression is positive, the critical point gives a minimum area.

For the maximum, check endpoints. Since $r$ must be non-negative and fit within the wire constraint, evaluate $r = 0$ and $r$ close to the boundaries of the wire.