A rectangle is inscribed in semicircle of radius 4m. Find the max possible area for such a rectangle

Results 1 to 5 of 5

- Jun 15th 2008, 06:54 AM #1

- Joined
- Jun 2008
- Posts
- 12

- Jun 15th 2008, 07:18 AM #2
Let $\displaystyle AB$ be the diameter, $\displaystyle O$ the center of the circle, $\displaystyle MNPQ$ the rectangle, with $\displaystyle M,Q\in[AB]$.

Let $\displaystyle PQ=x,OQ=y$. Then the area is $\displaystyle A=2xy$.

We have $\displaystyle x^2+y^2=16$.

If the area is maximum, then the squared area is also maximum.

$\displaystyle A^2=4x^2y^2=4x^2(16-x^2)$

Let $\displaystyle x^2=t$ and $\displaystyle f(t)=4t(16-t)=-4t^2+64t$

The maximum value of f is $\displaystyle \displaystyle f_{max}=-\frac{\Delta}{4a}$, where $\displaystyle \Delta=b^2-4ac, \ a=-4, \ b=64, \ c=0$.

So, $\displaystyle f_{max}=256$, for $\displaystyle t_{max}=-\frac{b}{2a}=8$.

Then $\displaystyle x=y=2\sqrt{2}$

- Jun 15th 2008, 08:04 AM #3

- Jun 15th 2008, 08:10 AM #4

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5

- Jun 15th 2008, 09:16 AM #5
Here is what I was doing until I realized I used the wrong radius.

Sorry if it seems redundant since RedDogs fine answer, but I worked it out earlier and do not want my attempt to go in vain.

Let x and y be the dimensions of the rectangle. Then area is**A=xy**

But $\displaystyle (\frac{x}{2})^{2}+y^{2}=16$

This means $\displaystyle y=\sqrt{16-\frac{x^{2}}{4}}=\frac{1}{2}\sqrt{64-x^{2}}$

Sub into A and we have:

$\displaystyle \frac{x}{2}\sqrt{64-x^{2}}$

$\displaystyle \frac{dA}{dx}=\frac{32-x^{2}}{\sqrt{64-x^{2}}}$

$\displaystyle 32-x^{2}=0, \;\ x=4\sqrt{2}, \;\ y=2\sqrt{2}$

Max area is then $\displaystyle (4\sqrt{2})(2\sqrt{2})=16$

Now, try the same problem with an inscribed trapezoid