Maximize area of rectangle inside a circle

[Feodalherren]

Homework Statement

Show that the maximum possible area for a rectangle inscribed in a circle is 2r^2 where r is the radius of the circle.

Homework Equations

Pre-calc

! NO TRIG !
Doesn't matter if it's easier, it's supposed to be solved with algebra.

The Attempt at a Solution

I have no clue what to do next or if those steps were even correct.

Hmm I just noticed that it's not correct. Whatever, I'm still stuck. Idk how to approach this problem.

 

[Bohrok]

I keep getting to a point where I can't solve it using just algebra and need to use calclus; still trying to figure out what else could be done...

Anyway, consider a sketch of the problem like this (but use r instead of 1) and see what you can do:

 

[Curious3141]

Homework Statement

Show that the maximum possible area for a rectangle inscribed in a circle is 2r^2 where r is the radius of the circle.

Homework Equations

Pre-calc

! NO TRIG !
Doesn't matter if it's easier, it's supposed to be solved with algebra.

The Attempt at a Solution

I have no clue what to do next or if those steps were even correct.

Hmm I just noticed that it's not correct. Whatever, I'm still stuck. Idk how to approach this problem.

Observe that any rectangle inscribed within a circle will have its centroid coincident with the centre of the circle. The centroid of the rectangle is the point at which its diagonals intersect. The length of a single diagonal is therefore equal to 2r.

Now let one side of the rectangle be x. Find the adjacent side using Pythagoras' theorem.

Set up an expression for the area A. You'll find there's a square root sign in it. Get rid of the square root sign by squaring both sides. You know have a relation between A² and x². Let z = x^2. Plot A² vs z - it's a simple parabola, and the horizontal (z) intercepts are easy to determine. Use symmetry to determine the z-value where the A² becomes a maximum. Observe that because of the monotonic nature of the square function for positive real numbers, the maximum of A² corresponds to the maximum of A. Hence find the maximum A value.

What "special" type of rectangle do you end up with?

 

[Feodalherren]

I don't get it. I still have two unknown values in my equation, namely r and x.

(2r)^2=x^2 + y^2 ? correct?

That would make y= sqrt(4r^2 - x^2)

A = (x)sqrt(4r^2 - x^2)

I'm missing something here.

 

[Curious3141]

I don't get it. I still have two unknown values in my equation, namely r and x.

(2r)^2=x^2 + y^2 ? correct?

That would make y= sqrt(4r^2 - x^2)

A = (x)sqrt(4r^2 - x^2)

I'm missing something here.

So far so good.

You can take r as a fixed quantity, since you're determining the rectangle with maximal area for a given circle.

Now work out A² as I advised, and put x² = z.

 

[eumyang]

This is probably an overkill method of doing this.

Show that the maximum possible area for a rectangle inscribed in a circle is 2r^2 where r is the radius of the circle.

Draw a circle in the Cartesian plane with center (0, 0). Inscribe a rectangle. We'll call the upper-right hand vertex of the rectangle (x, y). It should be straightforward that no matter what rectangle we choose with the corner at (x, y), the perimeter will be the same, namely P = 4x + 4y.

The area formula for this rectangle is, of course, A = 4xy.

Take the perimeter formula and solve for y. You'll have two variables on the other side, but treat P as a constant. Plug this expression into the area formula and you'll have a formula expressed in terms of x only (and the constant P). You will see that this is a quadratic.

The maximum/minimum value of a parabola is the y-coordinate of the vertex. In our case, the parabola will have a maximum value. (Do you see why?) Rewrite the area formula in vertex form. The x-coordinate of the vertex will be an expression in terms of P. The "y-coordinate" of this vertex is the maximum area. Take the original formula for the area, A = 4xy, and plug in the x-coordinate for x, and the "y-coordinate" for A. Solve for y, and you'll see right away the relationship between x and y.

See if you can get this far.

EDIT: You might as well stick with Curious3141's method. It's a little easier.

EDIT2: 1000th post.