Maximum area of a rectangle

Question

Two concentric circles have radii 13 and 15. Let ABCD be a rectangle, so that A and B lie on the larger circle, and C and D lie on the smaller circle. Find the maximum area of rectangle ABCD. I tried parametric equations but the number of variables were more than the number of equations. Is there a geometric solution possible. Thanks.

Answer 1

Let $AB=2x$, then $BC= \sqrt{169-x^2}+\sqrt{225-x^2}$. Do you know any calculus?

Answer 2

Let the radii of inner and outer circle be $r_1$ and $r_2$. Taking the central angle subtended by $CD$ to be $2\theta$ and by $AB$ to be $2\phi$ it is easily deduced with trigonometry that $AB=2r_1\sin\theta$ and $BC=(r_2\cos\phi-r_1\cos\theta)$. It's also easy to see that $\dfrac{r_2}{r_1}=\dfrac{\sin\theta}{\sin\phi}$ .

So the area of the rectangle is : $$area=2r_1\sin\theta(r_2\cos\phi-r_1\cos\theta)$$ $$area=2r_1^2\sin\theta\left( \sqrt{\frac{r_2^2}{r_1^2}-\sin^2\theta}-\cos\theta \right)$$ Differentiating this expression I found the derivative to be zero at $\tan\theta=\pm\dfrac{r_2}{r_1}$. Since the negative value gave the maximum, I figured that the $\cos\theta$ must be negative and then I substituted $\sin\theta=\dfrac{r_2}{\sqrt{r_1^2+r_2^2}}$ and $\cos\theta=\dfrac{-r_1}{\sqrt{r_1^2+r_2^2}}$ to get the maximal area$=\color{blue}{2r_1r_2}$

I'll think of a geometric solution later on, though I doubt I'll be able to think of one.(Geometry's not my forte)

Answer 3

One solution is that u assume the centre as origin and then by coordinate system assign the points of rectangle by symmetrical diagram.Then write the equation of circles and satisfy those points.Now differentiate the area wrt any one variable (side) .make sure that equation u got for area must be in same variable.

Answer 4

If $a,b$ are arms of a triangle, area formula $ A= \frac12 a\, b \sin C $ gives maximum area when the arms are perpendicular to each other. The relative angle $ C = 90^{0} $ between $a,b$ is important, sketched below in red.

From similar triangles,

$$ \sin \theta = \frac {h}{a} = \frac {b}{\sqrt{a^2+b^2}} ; $$

$$ h= \frac{ab}{\sqrt{a^2+b^2}}; $$

Area = $ 2 \cdot h \cdot {\sqrt{a^2+b^2}} = 2 \,a\, b = 2 \cdot13\cdot 15 = 390.$

Formula is $ A = 2 \,a\, b , $ nice problem set!