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A rectangle of largest area is inscribed in a semicircle of radius $r$. What is the area of the rectangle?

I just need the hint to solve it. How can I get length and breadth of rectangle in terms of radius $r$? If I can get length and breadth in terms of radius $r$, then I can solve $\frac{d (\text{Area})}{dr} = 0$.

enter image description here

Then $|AC| = r$.

Using pythagoras theorem, I will get

$$r = \left(b^2 + \frac{l^{2}}{4} \right)^{\frac{1}{2}}$$

Thanks in advance.

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Draw a picture and use the Pythagorean Theorem. Note that you would set one of the dimensions to be your variable. – David Mitra Mar 6 '12 at 4:35
@David i need to take two variables l and b, and find the value of l*b. – vikiiii Mar 6 '12 at 4:40
If you do what David says, you may see an equation relating $l$ and $b$. – Gerry Myerson Mar 6 '12 at 4:41
up vote 4 down vote accepted

It might be easier to deal with this using trigonometry.

Using your figure,

Notice that the area of the rectangle is four times the area of $\triangle{ABC}$.

Thus it is enough to maximize the area of the triangle.

Now if $\angle{BCA}$ is $\theta$, then the area of the triangle is $\frac{r^2}{2} \sin \theta \cos \theta = \frac{r^2}{4} \sin 2\theta$ (as $BC = r \cos \theta$ and $AB = r \sin \theta$).

Thus you need $\theta = \frac{\pi}{4}$ and the area of the rectangle is $r^2$

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Thanks @Aryabhata. This approach was the easiest. – vikiiii Mar 6 '12 at 5:10

The rectangle has upper corners on the semi-circle. Use this knowledge (and the knowledge about how to write an equation of the circle containing the semi-circle) to express the area $A=xy$ of the rectangle as a function of a single variable, then maximize it.

It will make life easier if you assume the center of the circle is at the origin.

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If you don't like doing things by halves, draw a rectangle inscribed in the semicircle. Then reflect the rectangle, and its semicircle, in the diameter of the semicircle. We get a rectangle inscribed in a circle of radius $r$. The old rectangle has maximum area if and only if the doubled rectangle has maximum area.

Let the sides of the doubled rectangle be $s$ and $t$. By the Pythagorean Theorem, $s^2+t^2=4r^2$.

The area of the doubled rectangle is $st$. We maximize this, or equivalently we maximize its square $s^2t^2$. But $s^2=4r^2-t^2$, so we maximize $t^2(4r^2-t^2)$, which is $4r^2t^2-t^4$. The usual calculus procedures tell us that the maximum is reached when $8r^2 t-4t^3=0$.

The solution $t=0$ is obviously not good for a maximum. But it is easy to verify that the solution $t=r\sqrt{2}$ gives a maximum. When $t=r\sqrt{2}$, we find that the maximum doubled rectangle is a square. Not a surprise!

So in the original semicircle, the inscribed rectangle of maximum area has base $r\sqrt{2}$ along the diameter, and height $r\sqrt{2}/2$.

Remark: To avoid calculus, which presumably you don't want to do since this is a calculus course, go back to the equation $s^2+t^2=4r^2$. We have $$s^2+t^2=(s-t)^2+2st,$$ and therefore $$2st=4r^2-(s-t)^2.$$ To maximize $st$, it is enough to maximize $2st$. To maximize $2st$, we minimize $(s-t)^2$. The minimum of $(s-t)^2$ is clearly $0$, reached when $s=t$, and therefore when $s=t=\sqrt{2}\,r$.

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