How do you maximize the perimeter of a rectangle inside a circle with equation: #x^2+y^2=1#?

Question

Emma Aldridge · Accepted Answer

#P=2sqrt2#

For symmetry reasons we can assume the rectangle has sides parallel to the axes. In such case if the corner in the first quadrant is the point #P(a,b)# with #0< a,b < 1#. The coordinates of the others are #P(+-a,+-b)# with the constraint:

#a^2+b^2 = 1#

so that:

#b= sqrt(1-a^2)#

and the perimeter is:

#P= 2a+2b = 2(a+sqrt(1-a^2))#

Evaluate the first derivative:

#(dP)/(da) = 2-(2a)/sqrt(1-a^2)#

And identify critical points solving the equation:

#(dP)/(da) = 0#

#1= a/sqrt(1-a^2)#

#a= sqrt(1-a^2)#

#a^2 = 1-a^2#

#a=1/sqrt2#

Evaluate the second derivative:

#(d^2P)/(da^2) = (-2sqrt(1-a^2)-2a/sqrt(1-a^2))/(1-a^2)#

#(d^2P)/(da^2) = (-2(1-a^2)-2a)/((1-a^2)sqrt(1-a^2))#

#(d^2P)/(da^2) = (-2-2a^2-2a)/((1-a^2)sqrt(1-a^2)) <0#

so that the critical point is a local maximum.

Then the perimeter is maximum when #a=b=1/sqrt2# and the rectangle is a square.

Adam Adkins · Answer

undefined To maximize the perimeter of a rectangle inside the circle with equation $x^2 + y^2 = 1$, the rectangle's vertices should lie on the circle. Let $P$ be the point on the circle where the rectangle's vertices meet. The rectangle's vertices are $(x, y)$, $(-x, y)$, $(-x, -y)$, and $(x, -y)$. The perimeter of the rectangle is $4x + 4y$.

Since the rectangle's vertices lie on the circle, they satisfy the equation of the circle: $x^2 + y^2 = 1$.

Using this information, we can maximize the perimeter using calculus:

1. Solve the equation of the circle for $y^2$: $y^2 = 1 - x^2$.
2. Substitute $y^2$ into the perimeter equation: $P = 4x + 4\sqrt{1 - x^2}$.
3. Differentiate $P$ with respect to $x$, and find critical points by setting the derivative equal to zero.
4. Solve for $x$ to find critical points.
5. Use the second derivative test to confirm the nature of the critical points.
6. Check the endpoints of the interval $[-1, 1]$ for maximum perimeter.

After these steps, you will find the value of $x$ that maximizes the perimeter, which you can then use to find the corresponding $y$ value.

Axel Alvarez · Answer

undefined To maximize the perimeter of a rectangle inside a circle with the equation x^2 + y^2 = 1, the rectangle must be inscribed in the circle, meaning its vertices lie on the circumference of the circle.

Let's denote the length of the rectangle as 2x and the width as 2y. The perimeter P of the rectangle is then given by P = 2(2x + 2y) = 4(x + y).

Since the rectangle is inscribed in the circle, its diagonal is the diameter of the circle, which has a length of 2. Using the Pythagorean theorem, we have (2x)^2 + (2y)^2 = 2^2, which simplifies to x^2 + y^2 = 1.

Now, we have two equations:
1. The perimeter equation: P = 4(x + y)
2. The constraint equation from the circle: x^2 + y^2 = 1

To maximize the perimeter, we need to optimize the expression 4(x + y) subject to the constraint x^2 + y^2 = 1.

We can solve this problem using the method of Lagrange multipliers or by expressing y in terms of x from the constraint equation and substituting it into the perimeter equation. After maximizing the resulting expression for P, we find the values of x and y that maximize the perimeter of the rectangle.

Once we find the optimal values of x and y, we can calculate the perimeter of the rectangle.