27 ft of wire is to be used to form an isosceles right triangle and a circle. Determine how much of the wire should be used for the circle if the total area enclosed is to be a minimum? Maximum?

if the triangle has side b, and the circle has circumference c, then

2b+b√2 + c = 27
so, b = (27-c)/(2+√2)

the area is

a = 1/2 b^2 + πr^2
= 1/2 ((27-c)/(2+√2))^2 + π(c/(2π))^2

This a parabola with vertex (minimum area) at

c = 27(1 - 1/(1+(3-2√2)π)) = 9.45

To determine how much of the wire should be used for the circle to minimize and maximize the total enclosed area, we need to understand the relationship between the wire length and the dimensions of each shape.

Let's start with the isosceles right triangle formed by the wire. An isosceles right triangle has two equal legs. Let's assume one leg has a length of 'x.' Since the other leg is also equal to 'x,' the hypotenuse can be found using the Pythagorean theorem: c = sqrt(2 * x^2) = x * sqrt(2).

Now, the circumference of the circle can be found using the remaining wire after forming the isosceles right triangle. The remaining wire length is 27 - 2x, as two legs of length 'x' are used for the triangle.

The circumference is given by the formula C = 2 * π * r, where 'r' is the radius of the circle. We need to find the value of 'r' that maximizes or minimizes the area.

Now, let's consider the total enclosed area, which consists of the area of the isosceles right triangle and the area of the circle.

1. To minimize the total enclosed area:
The area of the isosceles right triangle is given by A_triangle = (1/2) * base * height = (1/2) * x * x = (1/2) * x^2.
The area of the circle is given by A_circle = π * r^2.

The total area (A_total) is the sum of A_triangle and A_circle: A_total = A_triangle + A_circle = (1/2) * x^2 + π * r^2.

To minimize A_total, we need to minimize x and r. Since x is directly related to the wire length, we want to use the minimum allowed value. Thus, x = 0. The remaining wire length (27 - 2x) will be used for the circle.

To determine the radius, we use the formula for circumference: 2 * π * r = 27 - 2x. Substituting x = 0, we have 2 * π * r = 27, and solving for r gives r = (27 / (2 * π)).

2. To maximize the total enclosed area:
We need to maximize the isosceles right triangle's area, meaning we want to use the maximum allowed value for 'x.' In this case, x = (27 / 2), as we need to distribute the remaining wire equally on both legs.

The remaining wire length for the circle becomes 27 - 2x = 27 - 2 * (27 / 2) = 0.

Now, we can calculate the maximum value for the radius by using the formula for circumference: 2 * π * r = 0. Since the remaining wire is zero, the radius of the circle is also zero.

To summarize:
- To minimize the total enclosed area, use the remaining wire length (27 - 2x) for the circle, with a radius of (27 / (2 * π)).
- To maximize the total enclosed area, don't use any wire length for the circle, so the radius of the circle is zero.