calculusintegrationoptimization
find the area of the largest rectangle that can fit inside a semi circle of radius 2 cm
I have absolutely no idea where to get started on this…What I did do is
$A=(\pi r^2)/2$ (its a semi circle)
that gave me $6.28\ \mathrm{cm}^2$ now what do I do?
I
This is probably overkill, but if the polygon is convex and you fix the axis along which the rectangle is to be aligned, the problem is a second order cone program. If $x$ is a 2-d vector, any convex polygon can be expressed as the set of points satisfying $Ax \le b$, where A is a matrix and b is a vector. (Why? each bounding line is an inequality $a_i^T x \le b_i$, so then let $a_i^T$ be the ith row of $A$, and $b_i$ the ith component of b.)
So if $x$ is a corner of your rectangle, and $\ell,w$ are the length and width of your rectangle, the problem is: $$ \begin{eqnarray} \max_{x,\ell, w} \ \ell w \\ Ax \le b, \quad A\left(x + \left[\begin{matrix} 0 \\ w\end{matrix}\right] \right) \le b, \quad A\left(x + \left[\begin{matrix} \ell \\ 0\end{matrix}\right] \right)& \le b, \quad A\left(x + \left[\begin{matrix} \ell \\ w\end{matrix}\right] \right)& \le b, \quad \ell,w \ge 0 \end{eqnarray} $$ Then there's a standard trick to turn a hyperbolic constraint into an SOCP: $$\left \| \left[\begin{matrix} 2z \\ \ell - w\end{matrix}\right] \right\|_2 \le \ell + w \Leftrightarrow z^2 \le \ell w $$ So instead of maximizing $\ell w$ you can include the above constraint, and maximize $z$. If you don't have access to an SOCP solver is probably more complicated than what you need, but it does show how one can solve the problem exactly.
I think this image makes this problem much easier. By the Pythagorean Theorem we have $$r^2 + r^2 = (x\sqrt 2 + x + r)^2$$
Solving for $x$ we get:
$$r\sqrt 2 = x(\sqrt 2 + 1) + r$$
$$r(\sqrt 2 - 1) = x(\sqrt 2 + 1)$$
$$x = \frac{\sqrt 2 - 1}{\sqrt 2 + 1}r=\frac{(\sqrt 2 - 1)(\sqrt 2 - 1)}{(\sqrt 2 + 1)(\sqrt 2 - 1)}r=\frac{2-2\sqrt 2 + 1}{1}r=(3-2\sqrt 2)r$$
Best Answer
Let's agree that we only have to consider rectangles with two vertices on the diameter of the semicircle. This means that the four vertices have coordinates $\pm(r\cos\phi,0)$ and $\pm(r\cos\phi, r\sin\phi)$ for some $\phi\in\bigl[0,{\pi\over2}\bigr]$. The area of such a rectangle is $2r^2\cos\phi\sin\phi=r^2 \sin(2\phi)$, and this is maximal when $\phi={\pi\over4}$. Therefore the maximal possible area of such a rectangle is $r^2$.