## anonymous 5 years ago what is the maximum area of a square that can be fit in a circle of radius 'r'?

1. anonymous

Well what would the length of the diagonal line through the square be?

2. anonymous

if you assume the side of the square to be 'a' then the length of the diagonal will be $\sqrt{2}a$ and assume the rquation of the circle be $x ^{2}+y ^{2}=r ^{2}$

3. anonymous

*equation

4. anonymous

Yes, but if the square is inscribed in the circle, the points of the corners will lie on the circle. Therefore the diagonal to the square will be the diameter of the circle or twice the radius.

5. anonymous

So we have that $a\sqrt{2} = r \implies a = \frac{r}{\sqrt{2}}$ And you said that a was the length of the side, so using this we can find the area in terms of r. $$Area = a^2=\ ?$$

6. anonymous

you seem to have made a minor mistake.. shouldn't it be, $a \sqrt{2}=2r$ and so the area is inevitably equal to $a ^{2} = 2r^{2}$

7. anonymous

If the square has a side a then 2a^2 = r or a = sqrt(r/2) max area = sqrt(r/2)*sqrt(r/2) = r/2

8. anonymous

a^2 + a^2 = r^2 a = sqrt(r^2/2) Area = a*a = r^2/2

9. anonymous

one more time a^2 + a^2 = 4r^2 a = sqrt(r^2/2) Area = a*a = 4r^2/2

10. anonymous

last time a^2 + a^2 = 4r^2 a = sqrt(4r^2/2) Area = a*a = 4r^2/2

11. anonymous

or Area = 2r^2

12. anonymous

Yes sorry, that's correct.

13. anonymous

good work ivan.. but using polpak's idea to find it out using the diagonal of the square is a hell lot a shorter

14. anonymous

$a\sqrt{2} = 2r \implies a = \frac{2r}{\sqrt{2}} \implies a^2 = \frac{4r^2}{2} = 2r^2$

15. anonymous

This also can be interesting. The area of a square with a side x is $A(x) = x^{2}$ The distance of any two points inside of a circle must be less than or equal to 2r The longest distance in a square is the diagonal $diagonal = \sqrt{x^{2} + x ^{2}} = \sqrt{2x ^{2}}$ so, we have: $\sqrt{2x ^{2}}\le2r$ or $x \le \pm \sqrt{2r ^{2}}$ $A(x) = x ^{2}, D:\left\{ x | x \le \pm \sqrt{2r ^{2}} \right\}$ $A \prime(x) = 2x, D=\left\{ x | x <\pm\sqrt{2r ^{2}} \right\}$ $A \prime \prime(x) = 2, D=\left\{ x | x <\pm\sqrt{2r ^{2}} \right\}$ $A \prime(x) = 0, for x = 0$ $A \prime(x) > 0 , for x > 0$ $A \prime(x) < 0 , for x < 0$ $A \prime \prime(x) > 0, for all x$ => A(x) is concave up for all x $=> A(x) is decreasing for-\sqrt{2r ^{2}} < x <0$ $=> A(x) is increasing for: 0 < x < \sqrt{2r ^{2}}$ => A(x) has a minimum at x = 0, A(0) = 0 => A(x) desn't have a maximum value in the open interval $(-\sqrt{2r ^{2}},\sqrt{2r ^{2}})$ => the maximum value of A(x) on the closed interval $[-\sqrt{2r ^{2}}, \sqrt{2r ^{2}}]$must be at$x \in \left\{ -\sqrt{2r ^{2}},\sqrt{2r ^{2}} \right\}$ => the maximum area of a square inside a circle is: $A(\pm\sqrt{2r ^{2}}) =(\pm\sqrt{2r ^{2}})^{2} = 2r ^{2}$

16. anonymous

boy! was that scary! lol