1. ## Optimization Problem

Without using methods from calculus, find the optimal dimensions $x$ and $y$, representing length and width, for a rectangle that must be contained within a unit semi-circle.

P.S: Rotate the rectangle by 90 degrees and re-examine the problem

2. Originally Posted by Deco
Without using methods from calculus, find the optimal dimensions $x$ and $y$, representing length and width, for a rectangle that must be contained within a unit semi-circle.
Start by writing the area in terms of the width of the rectangle. Since the rectangle fits in the semi-unit circle, we have if the width is $w$, its height is: $h=\sqrt{1-(w/2)^2}$.

Better yet start by drawing a picture.

CB

3. I need actual numbers, but yes, you got the jist of it.

4. Originally Posted by Deco
Without using methods from calculus, find the optimal dimensions $x$ and $y$, representing length and width, for a rectangle that must be contained within a unit semi-circle.

P.S: Rotate the rectangle by 90 degrees and re-examine the problem
Could be that my reasining is wrong but nevertheless I'l give it a try:

Since Archimedes it is known that a square is the optimal rectangle inscribed in a circle. The side length of this square is $s = r\cdot \sqrt{2}$.

Now cut this figure parallel to one side of the square into 2 congruent parts: The circle becomes a semi-circle and the sqare is now a rectangle with the original side of the square as length and half of the length as height.

With your problem you have $r = 1$.

5. Originally Posted by Deco
I need actual numbers, but yes, you got the jist of it.
And what have you done (my post was a hint to get you started)?

CB

6. Originally Posted by earboth
Could be that my reasining is wrong but nevertheless I'l give it a try:

Since Archimedes it is known that a square is the optimal rectangle inscribed in a circle. The side length of this square is $s = r\cdot \sqrt{2}$.

Now cut this figure parallel to one side of the square into 2 congruent parts: The circle becomes a semi-circle and the sqare is now a rectangle with the original side of the square as length and half of the length as height.

With your problem you have $r = 1$.
But now they will have to reproduce Archimedes proof for the square (which can be done in the same way as the direct approach to this by competeing the square (no pun intended) of a quadratic derived from the area formula).

CB

7. I've already completed this problem, I was putting it up as a challenge.
Assuming a rectanle that is inscribed with a semi cirlce has two of it's
corners intersecting (roughly speaking) the curve of the circle.

Since we want both width, $w$, and height, $h$
must be at an optimal point, allowing for the maximum area.

We can find:

$w \times h= A$

$w = 2 \cos \theta$

The above is true because of the geometry of the problem.
If rotated 90 degrees h would be exactly one-half the total width.

and,

$h = \sin \theta$

Resubbing, we get:

$(2 \cos \theta ) ( \sin \theta ) = A$

$\sin 2 \theta = 1$

Above is true because oscillations in the sine function make
the maximum value of it to be 1.

$2 \theta = \sin^ -1 (1)$

$2 \theta = 90^\circ$

$\theta = 45^\circ$

Therefore,
$\sin 45^\circ = \frac {\sqrt{2}}{2}$

$h = \frac {\sqrt{2}}{2}$

$2 \cos 45^\circ = \sqrt {2}$

$w= \sqrt{2}$

...I think