EXAMPLE $5$ Find the area of the largest rectangle that can be inscribed in a semicircle of radius $r.$

SOLUTION $1$ Let's take the semicircle to be the upper half of the circle $x^2+y^2=r^2$ with center the origin. Then the word inscribed means that the rectangle has two vertices on the semicircle and two vertices on the $x$ axis as shown in the top figure.

Let $(x,y)$ be the vertex that lies in the first quadrant. Then the rectangle has sides of lengths $2x$ and $y,$ so its area is $A=.$

To eliminate $y$ we use the fact that $(x,y)$ lies on the circle $x^2+y^2=r^2$ and so $y=.$ Thus $A=.$

The domain of this function is $0xr.$ Its derivative is

$A^{\text{'}}=2\sqrt[\phantom{2}]{r^2-x^2}-\frac{2x^2}{\sqrt[\phantom{2}]{r^2-x^2}}=\frac{2(r^2-2x^2)}{\sqrt[\phantom{2}]{r^2-x^2}}$

which is $0$ when $2x^2=r^2,$ that is$,x=q,($since $x0).$ This value of $x$ gives a maximum value of A since $A(0)=0$ and $A(r)=0.$ Therefore the area of the largest inscribed rectangle is

$A(\frac{r}{\sqrt[\phantom{2}]{2}})=2\frac{r}{\sqrt[\phantom{2}]{2}}$

$q,.$

SOLUTION $2$ A simpler solution is possible if we think of using an angle as a variable. Let $$ be the angle shown in the bottom figure. Then the area of the rectangle is

$A()=(2rcos())(rsin())=r^2(2sin()cos())=r^{2}sin(2).$

We know that $sin(2)$ has a maximum value of $1$ and it occurs when $2=\frac{}{2}.$ So $A()$ has a maximum value of $r^2$ and it occurs when $=\frac{}{4}.$

Notice that this trigonometric solution doesn't involve differentiation. In fact, we didn't need to use calculus at all.

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