a pc of wire 10feet long is cut into 2 pcs. one piece is bent into the shape of a circle and the other into the shape of a square. how should the wire be cut so that the combined area is as large as possible?
Intuitively, the circle has a bigger area for the same circumference, so we try to make a bigger circle.
Also, since the area increases as the square of the circumference, we want to make 1 single shape as opposed to two.
So intuitive answer is to make the smallest square possible (zero if allowed).
Let's check our intuition.
Let x=length to make the circle
Radius of circle, r
=x/(2π)
Area of circle
=πr²
=x²/(4π)
For the square,
perimeter=10-x
side length, s
=(10-x)/4
Area of square
=((10-x)/4)²
Total area, A(x)
=x^2/(4*π)+((10-x)/4)²
dA(x)/dx=x/(2π)-(10-x)/8
For maximum, equate dA(x)/dx to zero, and solve for x.(=4.399)
Check if the answer is a maximum or minimum by graphing, or calculate d2A(x)/dx².
If d2A(x)/dx².>0, it is a minimum, and thus has no use for us.
In that case, we calculate A(0), A(4.399) and A(10) to find the areas at the critical points and choose the largest one.
This is a very common question in introductory Calculus, and each time I have seen it, it asked for a MINIMUM area , and as MathMate noted, we would make the circle as large as possible.
Taking it with the usual minimum area request,
let the radius of the circle be r
let the side of the square be x
then 2πr + 4x = 10
x = (5 - πr)/2
A = πr^2 + x^2 = πr^2 + (1/4) (25 - 10πr + π^2 r^2)
dA/dr = 2πr + (1/4)(-10 +2π^2 r) = 0 for a min of A
dividing by π and solving for r,
r = 5/(4+π) = appr .7001
x = 1.4002
since 4x is needed for the square, the wire should be cut
appr 5.6 cm for the square and 4.4 cm for the circle.
To find the best way to cut the wire so that the combined area is as large as possible, we can use optimization techniques. Let's break down the problem step by step:
1. Understand the problem:
The given wire of length 10 feet needs to be cut into two pieces to form a circle and a square. We want to maximize the combined area formed by these shapes.
2. Identify variables:
Let's denote the length of the wire used for the circle as 'x' feet and the length used for the square as 'y' feet. Since the total length of the wire is 10 feet, we have the equation: x + y = 10.
3. Define equations for area:
The area of a circle is given by A_circle = π * r^2, where r is the radius.
The area of a square is given by A_square = side^2, where side is the length of one side of the square.
4. Relate variables and create an equation for the combined area:
We can write the total area, A_total, as the sum of the areas of the circle and the square: A_total = A_circle + A_square.
5. Express one variable in terms of the other:
Since we have two variables, x and y, we need to express one variable in terms of the other. We will express y in terms of x from the equation x + y = 10. Therefore, y = 10 - x.
6. Substitute the expression for one variable into the equation for the combined area:
Substituting the expression for y into the equation A_total = A_circle + A_square, we get:
A_total = π * r^2 + (10 - x)^2.
7. Maximize the combined area:
Now, we need to find the value of x that maximizes the combined area. To do this, we take the derivative of the combined area equation with respect to x, set it equal to zero, and solve for x. We can then substitute this value back into the equation to find the corresponding y.
8. Calculate the solution:
Taking the derivative of A_total = π * r^2 + (10 - x)^2 with respect to x, we get:
dA_total/dx = -2(10 - x).
Setting dA_total/dx equal to zero:
-2(10 - x) = 0
20 - 2x = 0
2x = 20
x = 10.
Substituting x = 10 back into the equation y = 10 - x, we find:
y = 10 - 10
y = 0.
9. Evaluate the solution:
We have x = 10 and y = 0. This means that the wire should be cut into a length of 10 feet for the circle and 0 feet for the square. The reason for this is that a circle has the largest area among all shapes with a given perimeter, while a square has the smallest area among all shapes with a given perimeter. Hence, having as much wire as possible for the circle and none for the square will maximize the combined area.
Therefore, to maximize the combined area, the wire should be cut such that the entire 10 feet is used to form the circle, and no wire is used for the square.