A piece of wire 25 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 of the wire should go to the square to maximize the total area enclosed by both figures?
(b) how much of the wire should go to the square to minimize the total area enclosed by both figures?
Alternate solution
b)
Let the radius of the circle be r
the we need 2πr for the circle
leaving 25-2πr for the square, then each side of the square is
(25-2πr)/4
Area = πr^2 + (25-2πr)^2 /16
= πr^2 + 625/16 - (100/16)πr + (4π^2)/16 r^2
this will graph as a parabola opening upwards, so it will have a minimum area.
d(area)/dr = 2πr - 100π/16 + π^2 r/2
= 0 for a max of area
2πr + π^2 r/2 = 100π/16
times 16 , then solving for r
32πr + 8π^2 r = 100π
r(32π + 8π^2) = 100π
r = 100/(32+8π) = 25/(8+2π) appr. 1.7503
then the circle needs 2πr = appr. 10.998
leaving 25 – 10.998 or 14.003 m for the square.
Plugging r = 1.7503 into my area equation gives
Minimum Area = 21.879
test by taking a value of r slightly above and slightly below r = 1.7503
if r = 1.7 , area = 21.893
if r = 1.7503 , area = 21.879
if r = 1.8 , area = 21.893
a) The maximum area is obtained when all the wire is used for the circle and NONE is used for the square
i.e. , max area = (12.5^2)π = 490.87
how is the derivative equal to that? shouldn't the last part (ð^2 r/2)be pi*2r/2?
(a) To maximize the total area enclosed by both figures, we need to find the optimal distribution of wire between the square and the circle.
Let's assume x meters of the wire will be used to create the square. Since the wire cut into two pieces, the remaining wire used for the circle will be (25 - x) meters.
For the square, each side will have a length of (x/4) meters, so the total area of the square will be (x/4) * (x/4) = (x^2 / 16) square meters.
For the circle, the circumference will be (25 - x) meters. The radius of the circle will be [(25 - x) / (2π)] meters. Thus, the area of the circle will be π * [(25 - x) / (2π)]^2 = (25 - x)^2 / (4π) square meters.
The total area enclosed by both figures will be the sum of the areas of the square and the circle, which is A = (x^2 / 16) + (25 - x)^2 / (4π).
To find the value of x that maximizes A, we can take the derivative of A with respect to x and set it equal to zero:
dA/dx = (2x / 16) - (2(25 - x) / (4π)) = 0.
Simplifying, we get x / 8 - (25 - x) / (2π) = 0.
Multiplying through by 8π, we get xπ - (200 - 8x) = 0.
Rearranging the equation, we get (xπ + 8x) = 200.
Factoring out x, we get x(π + 8) = 200.
Dividing both sides by (π + 8), we get x = 200 / (π + 8).
So, to maximize the total area enclosed by both figures, approximately 8.370 meters of wire should go to the square.
(b) To minimize the total area enclosed by both figures, we need to find the optimal distribution of wire between the square and the circle.
Following the same logic as above, we know that x meters of wire will be used to create the square, while (25 - x) meters will be used for the circle.
The total area enclosed by both figures will be A = (x^2 / 16) + (25 - x)^2 / (4π).
To find the value of x that minimizes A, we can again take the derivative of A with respect to x and set it equal to zero:
dA/dx = (2x / 16) - (2(25 - x) / (4π)).
Setting this equal to zero and following the same steps as above, we find that x = 200 / (π - 8).
So, to minimize the total area enclosed by both figures, approximately 8.184 meters of wire should go to the square.
To find the answer to this question, we need to apply some principles of geometry. Let's break down the problem step by step:
(a) To maximize the total area enclosed by both figures, we need to figure out the optimal proportions of the wire to be allocated to each shape.
Let's assume x meters of the wire is used for the square. This means the remaining 25 - x meters will be used for the circle.
For the square:
The perimeter of a square is given by 4 times the length of one of its sides.
Since the wire allocated to the square is x meters, the length of each side will be x/4.
The area of the square is given by the square of the length of a side, so it will be (x/4)^2.
For the circle:
The perimeter of a circle is given by the formula 2 * π * r, where r is the radius.
We know that the remaining wire allocated to the circle is 25 - x meters, and this length corresponds to the circumference of the circle, so we can write the equation as:
25 - x = 2 * π * r
To find the area of the circle, we need to know the radius. The radius can be found by dividing the circumference equation by 2 * π:
r = (25 - x) / (2 * π)
The area of a circle is given by the formula π * r^2, so the area will be given by:
π * ((25 - x) / (2 * π))^2
Now, to find the total area enclosed by both figures, we need to add the area of the square to the area of the circle:
Total Area = (x/4)^2 + (π * ((25 - x) / (2 * π))^2)
Simplifying this equation:
Total Area = (x^2)/16 + ((25 - x)^2) / 4
To maximize the total area, we need to find the value of x that maximizes this equation. We can do this by taking the derivative of the equation with respect to x, setting it equal to zero, and solving for x.
(b) To minimize the total area enclosed by both figures, we follow the same steps as above but minimize the equation instead of maximizing it.
let the length of side of the square be x
Now, the area of a circle is pi r^2, but we have a circumference c = 2pi*r
r = c/2pi and c = (25-4x)
a = x^2 + pi * [(25-4x)/2pi]^2
= x^2 + (25-4x)^2/4pi
This function has a minimum at x = 25/(4 + pi) = 3.5
area is 21.88
but has no global maximum.
At x=0 the area is just the circle = pi * (25/2pi)^2 = 625/4pi = 49.73
At x= 25/4 the area is just the square = 625/16 = 39.06