The combined perimeter of a circle and a square is 16. Find the dimension of the circle and the square that produce a maximum total area?
since the coefficient of r^2 is positive, A has a minimum at the vertex, not a maximum.
To maximize the total area, we need to determine the dimensions of the circle and square that satisfy the given perimeter constraint.
Let's assume the side length of the square is "s" and the radius of the circle is "r".
The formulas to calculate the perimeter of the square and the circumference of the circle are as follows:
Perimeter of the square = 4s
Circumference of the circle = 2πr
According to the given information, the combined perimeter of the square and circle is 16. Therefore, we have:
4s + 2πr = 16 (Equation 1)
To find the dimensions that maximize the total area, we need to express the area of the square and the area of the circle in terms of "s" and "r" respectively.
Area of the square = s^2
Area of the circle = πr^2
The total area is the sum of the area of the square and the area of the circle:
Total Area = s^2 + πr^2
We need to find the maximum value of "s" and "r" that satisfies Equation 1 and maximizes the Total Area.
To solve for the values of "s" and "r", we will use the method of Lagrange multipliers.
1. Set up the Lagrange function:
L(s, r, λ) = s^2 + πr^2 + λ(4s + 2πr - 16)
2. Take partial derivatives of L(s, r, λ) with respect to s, r, and λ:
∂L/∂s = 2s + 4λ = 0 (Equation 2)
∂L/∂r = 2πr + 2πλ = 0 (Equation 3)
∂L/∂λ = 4s + 2πr - 16 = 0 (Equation 4)
3. Solve the system of equations (Equations 2, 3, and 4) simultaneously.
From Equation 2, we get:
2s + 4λ = 0
s = -2λ (Equation 5)
From Equation 3, we get:
2πr + 2πλ = 0
r = -λ (Equation 6)
Substituting Equations 5 and 6 into Equation 4, we get:
4(-2λ) + 2π(-λ) - 16 = 0
-8λ - 2πλ - 16 = 0
λ(8 + 2π) = -16
λ = -16 / (8 + 2π)
λ = -8 / (4 + π) (Equation 7)
Substituting the value of λ from Equation 7 into Equations 5 and 6, we can solve for the dimensions:
s = -2λ
s = -2(-8 / (4 + π))
s = 16 / (4 + π)
s = 16 / (4 + 3.14) (Substituting π ≈ 3.14)
s = 16 / 7.14
s ≈ 2.238
r = -λ
r = -(-8 / (4 + π))
r = 8 / (4 + π)
r = 8 / (4 + 3.14) (Substituting π ≈ 3.14)
r = 8 / 7.14
r ≈ 1.12
Therefore, the approximate dimensions that produce a maximum total area are:
- Side length of the square ≈ 2.238
- Radius of the circle ≈ 1.12
Let the side length of the square be $x$, and let the radius of the circle be $r$. Then the perimeter of the square is $4x$ and the perimeter of the circle is $2\pi r$. We are given that their sum is 16, so we have the equation $4x+2\pi r=16$.
We want to find the dimensions that produce a maximum total area, which consists of the area of the square ($x^2$) and the area of the circle ($\pi r^2$). We can write this as:
$A=x^2+\pi r^2$
We can use the equation we derived earlier to solve for $x$ in terms of $r$: $x=\frac{16-2\pi r}{4}=4-\frac{\pi}{2}r$.
Substitute this expression for $x$ into the equation for $A$ to get:
$A=\left(4-\frac{\pi}{2}r\right)^2+\pi r^2$
Expanding the square and simplifying, we get:
$A=16-4\pi r+r^2$
This is a quadratic function in $r$ with a maximum value at the vertex. The $r$-coordinate of the vertex is given by $-\frac{b}{2a}=-\frac{-4\pi}{2}=2\pi$. Therefore, the radius of the circle that maximizes the total area is $r=2\pi$.
Substitute this value of $r$ into the equation for $x$ to get:
$x=4-\frac{\pi}{2}(2\pi)=4-\pi^2$
Therefore, the dimensions of the circle and the square that produce a maximum total area are:
Radius of circle: $2\pi$
Side length of square: $4-\pi^2$
You are correct, I apologize for the error. Since the coefficient of $r^2$ is positive, the equation for $A$ has a minimum value at the vertex. Therefore, the dimensions of the square and the circle that produce a minimum combined area are:
Radius of circle: $2\pi$
Side length of square: $4 - \pi^2$
Thank you for catching that mistake.