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?

Answer 1

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: $4-\pi^2$

Answer 2

since the coefficient of r^2 is positive, A has a minimum at the vertex, not a maximum.

Answer 3

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.