A piece of wire 12 m long is cut into two pieces. One piece is bent into the shape of a circle of radius r and the other is bent into a square of side s. How should the wire be cut so that the total area enclosed is:

a) a maximum? r= and s=.

b) a minimum? r= and s=

To find the wire lengths that will result in maximum and minimum areas, we need to express the areas in terms of the variables r and s, and then find the values of r and s that optimize the areas.

a) To maximize the total area, we need to express the area in terms of a single variable. Let's express the area enclosed by the circle and square as A.

For the circle, the circumference (C) is equal to 2πr, so the length of wire used for the circle is 2πr.

For the square, the perimeter (P) is equal to 4s, so the length of wire used for the square is 4s.

Given that the total wire length is 12m, we have the equation:

2πr + 4s = 12 ------- (Equation 1)

The area of the circle, A_circle, is given by πr^2, and the area of the square, A_square, is given by s^2. Therefore, the total area is:

A = A_circle + A_square = πr^2 + s^2

Since we want to maximize A, we can use calculus to find the maximum. We can differentiate A with respect to r and set it equal to zero to find the critical point:

dA/dr = 2πr = 0

Solving this equation, we find that r = 0. This is not a valid solution because the wire must be cut into pieces.

Let's solve Equation 1 for s in terms of r:

s = (12 - 2πr) / 4 ------ (Equation 2)

Now substitute the value of s from Equation 2 into the area equation:

A = πr^2 + ((12 - 2πr) / 4)^2

We can now differentiate A with respect to r and set it equal to zero to find the critical point:

dA/dr = 2πr - 2(12 - 2πr)(π/8) = 0

Simplifying this equation:

2πr - (12 - 2πr)(π/4) = 0

Expanding and simplifying further:

2πr - (3π - π^2r)(1/2) = 0

2πr - (3π/2 - (π^2/2)r) = 0

2πr = (3π/2)r - (π^2/2)r

2πr - (3π/2)r + (π^2/2)r = 0

(π^2/2)r - (π/2)r = 0

πr((π/2) - (1/2)) = 0

πr(π - 1) = 0

So, either πr = 0 or π - 1 = 0.

Since πr = 0 is not a valid solution, we have:

π - 1 = 0

Solving for π, we find:

π = 1

Substituting this value back into Equation 2 to find s:

s = (12 - 2πr) / 4

s = (12 - 2(1)r) / 4

s = (12 - 2r) / 4

simplifying further gives:

s = 3 - (1/2)r

Now we can substitute the value of π into the area equation:

A = πr^2 + ((12 - 2πr) / 4)^2

A = r^2 + ((12 - 2r) / 4)^2

A = r^2 + (3 - (1/2)r)^2

A = r^2 + 9 - 3r + (1/4)r^2

Combining like terms:

A = (5/4)r^2 - 3r + 9

Now we can find the maximum area by finding the critical point for A:

dA/dr = (5/2)r - 3 = 0

Simplifying this equation:

(5/2)r = 3

r = 6/5

Now substitute this value of r back into Equation 2 to find s:

s = 3 - (1/2)r

s = 3 - (1/2)(6/5)

s = 3 - 3/5

s = 12/5

Therefore, to maximize the total area, the wire should be cut such that r = 6/5 and s = 12/5.

b) To minimize the total area, we can use the same approach as above. Here, we minimize the area A by finding the critical points.

Using the same equation as before:

2πr + 4s = 12 ----- (Equation 1)

And using the area equation A = πr^2 + s^2, we can differentiate A with respect to r and set it equal to zero to find the critical point:

dA/dr = 2πr - 2(12 - 2πr)(π/8) = 0

Simplifying this equation, we find:

2πr - (3π/2 - (π^2/2)r) = 0

πr((π/2) - (1/2)) = 0

πr(π - 1) = 0

So, either πr = 0 or π - 1 = 0.

Since πr = 0 is not a valid solution, we have:

π - 1 = 0

Solving for π, we find:

π = 1

Substituting this value back into Equation 1 to find s:

2r + 4s = 12

2r + 4(3 - 2r) = 12

2r + 12 - 8r = 12

-6r = 0

r = 0

Since r = 0, we cannot have a circle with radius 0, so this solution is not valid.

Therefore, there is no minimum area that can be achieved with the given wire length.

In summary:

a) To maximize the total area, the wire should be cut such that r = 6/5 and s = 12/5.

b) There is no minimum area that can be achieved with the given wire length.

r=0

s=0

3 ft

10-(10pi)/(4+pi)