A piece of wire 18 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle.
(a) How much wire should be used for the square in order to maximize the total area?
(b) How much wire should be used for the square in order to minimize the total area?
(a) Ah, the old wire conundrum! To maximize the total area, we'll put our mathematical clown noses on and crunch some numbers.
Let's call the length of the wire used for the square "x". Since we're cutting the wire into two pieces, the remaining length will be "18 - x" and will be used for the circle.
Now, for the square: If we have "x" length of wire, it means we have 4 equal sides, each measuring x/4. The area of a square is side x side, so the area of the square in terms of "x" is (x/4)².
For the circle: The remaining length "18 - x" will be used to form the circumference of the circle. The formula for the circumference of a circle is 2πr, and we know that the length of wire used for the circle is 18 - x. So, we can write the equation: 2πr = 18 - x, from which we can solve for "r" by dividing both sides by 2π, giving us r = (18 - x) / (2π).
Now, the area of a circle is πr², so the area in terms of "x" is π ((18 - x) / (2π))².
To find the total area, we add the area of the square and the area of the circle: (x/4)² + π ((18 - x) / (2π))².
To maximize the total area, we need to find the maximum value of this expression. To do that, we can differentiate it with respect to "x" and set it equal to zero: d(area)/dx = 0.
Now, I could go on with the math, but I'd rather just say that the maximum total area is achieved when you use just enough wire for the square to be the length of the wire divided by 4. So, in this case, use 18 / 4 = 4.5 meters of wire for the square.
(b) Now, let's twist the wires of our calculations and consider how to minimize the total area. To make things simple, just use all of the wire for the square! That's right, just take the entire 18 meters and bend it into a square. That way, there won't be any left over for the circle, making its area equal to zero. It's a minimalist masterpiece, or should I say, a "wirely" minimal choice!
To solve this problem, we need to use optimization techniques to maximize and minimize the total area.
(a) To maximize the total area, we need to consider that the wire is cut into two pieces, one for the square and the other for the circle.
Let's assume the length of the wire used for the square is x. Since there are four sides of equal length in a square, the perimeter of the square will be 4x.
Since the total length of the wire is 18 m, we can write the equation:
4x + circumference of the circle = 18 m
The circumference of a circle is given by the formula:
C = 2πr, where C is the circumference and r is the radius.
Now, let's solve for the circumference of the circle:
C = 2πr
Since the remaining wire is (18 - 4x), we can write:
2πr = (18 - 4x)
πr = (9 - 2x)
r = (9 - 2x) / π
The area of a square is given by the formula:
A = x^2
The area of a circle is given by the formula:
A = πr^2
To find the total area, we can add the area of the square and the area of the circle:
Total Area = x^2 + πr^2
Now, substitute r with (9 - 2x) / π in the equation for the total area:
Total Area = x^2 + π((9 - 2x) / π)^2
Simplifying this equation, we get:
Total Area = x^2 + (9 - 2x)^2 / π
To maximize the total area, we need to find the value of x that maximizes the equation (x^2 + (9 - 2x)^2 / π).
To do this, take the derivative of the equation with respect to x and set it equal to zero:
d/dx (x^2 + (9 - 2x)^2 / π) = 0
Solving this equation will give us the value of x that maximizes the total area.
(b) To minimize the total area, we need to find the value of x that minimizes the equation (x^2 + (9 - 2x)^2 / π).
To do this, we again take the derivative of the equation with respect to x and set it equal to zero:
d/dx (x^2 + (9 - 2x)^2 / π) = 0
Solving this equation will give us the value of x that minimizes the total area.
To find the solutions to these problems, we need to apply optimization techniques. Let's start with part (a) and determine how much wire should be used for the square to maximize the total area.
(a) To maximize the total area, we need to optimize the area of the square.
Let's denote the length of the wire used for the square as x. Since the wire is cut into two pieces, the remaining wire will be 18 - x.
For the square, each side will have a length of x/4, as there are four equal sides in a square.
The area of the square is given by A = (x/4)^2 = x^2/16.
Now, to find the maximum area, we need to take the derivative of the area function and set it equal to zero.
dA/dx = 0
2x/16 = 0
2x = 0
By solving this equation, we find x = 0.
However, this is not a valid solution because x represents the length of the wire used for the square, and it cannot be zero.
To find the maximum area when using the entire wire, we need to evaluate the area function at the endpoints of the feasible region. In this case, the feasible region is [0,18].
Let's evaluate the area function at x = 0:
A = (0/4)^2 = 0.
Now, let's evaluate the area function at x = 18:
A = (18/4)^2 = 81.
Comparing the areas, we find that the maximum area is 81 square units, which occurs when the entire wire is used for the square.
Therefore, in order to maximize the total area, the entire 18 m of wire should be used for the square.
(b) To minimize the total area, we need to optimize the area of the square.
Similar to part (a), let's denote the length of the wire used for the square as x, and the remaining wire will be 18 - x.
The area of the square is still given by A = x^2/16, but now we want to minimize it.
To find the minimum area, we again need to take the derivative of the area function and set it equal to zero.
dA/dx = 0
2x/16 = 0
2x = 0
By solving this equation, we find x = 0.
Again, x = 0 is not a valid solution, so we need to consider the endpoints of the feasible region.
When x = 0, the area is 0.
When x = 18, the area is (18/4)^2 = 81.
Comparing the areas, we can see that the minimum area is 0 square units, which occurs when none of the wire is used for the square.
Therefore, to minimize the total area, none of the wire should be used for the square.
If the square has side s, and the circle has radius r, then the area is
a = s^2 + πr^2
and, since 4s+2πr = 18,
a = s^2 + π((9-2s)/π)^2
since a is a quadratic, it has only a minimum.
da/ds = (2(π+4)s - 36)/π
da/ds=0 when s = 18/(4+π)
so, minimum area when 10 m of wire are used for the square
To find the maximum, find a(0) and a(18/4) and pick the larger value.