Question

Calculus: what is the largest possible area of a rectangle inscribed in a circle with a radius of r.

Answer 1

Here is my work. I came up with a very odd answer though, any ideas?

Answer 2

here is a non calculus method to check

if you draw perpendicular diameters in the circle, then the endpoints will form the vertices of the square

a square is a specific rhombus

the area of a rhombus can be found using the formula

\[A = \frac{1}{2} xy\]

where x and y are the diagonals of the rhombus

in the question you have, the diagonals will both be 2r..(diameters of the circle)

so the area will be

\[A = \frac{1}{2} \times 2r \times 2r\]

hope it agrees to your answer

Answer 3

and looking at your solution, if you differentiate with respect to x

you have a mistake in the product rule

\[v = (r^2 - x^2)^{\frac{1}{2}}~~~~~v' = \frac{1}{2} (r^2 - x^2)^{-\frac{1}{2}} \times -2x\]

Answer 4

does the setup look correct? the formulas at least?

Answer 5

wouldn't it be this? \[v = (r^2 - x^2)^{\frac{1}{2}}~~~~~v' = \frac{1}{2} (r^2 - x^2)^{-\frac{1}{2}} \times (2r -2x)\]

Answer 6

No, campbell_st's version is correct

derivative of \(r^2\) with respect to \(x\) is \(0\)

Answer 7

oh, gotcha, but the original formulas look ok, just the math mistake?

Answer 8

yes, looks like it.

Answer 9

I came up with x=sqrt(2r^2) , can someone verify that this is correct?

Answer 10

Well, let's just take it as obvious that a square with diagonal length \(2r\) is going to be the biggest rectangle you can stuff in there, by a symmetry argument. The answer can't change as you rotate the circle or rectangle, right?

That means our situation looks like this:



\(a = b\) by symmetry, so we have by Pythagorean theorem \[a^2+a^2=r^2\]\[2a^2=r^2\]\[a=\frac{r}{\sqrt{2}}\]
I'm not sure what you are referring to by \(x\) in your question, but hopefully you have enough illumination to see your way to the area of the inscribed rectangle from here.