Maximize the area of a triangle inscribed in a semicircle.

Question

https://i.stack.imgur.com/YZ679.png

Consider the semicircle with radius 1, the diameter is AB. Let C be a point on the semicircle and D the projection of C onto AB. Maximize the area of the triangle BDC.

My attempt so far, I'm new at these problems and have only done a few so far. Thanks for any hints. I feel I'm not setting up the correct equation.

$x^{2}$+$y^2$=$r^2$

$y=\sqrt{r^2 - x^2}=\sqrt{1-x^2}\;(\text{since}\:r=1).$

The area is A=$\dfrac{1}{2}$($x+1$)($y$)

A'=$\frac{2x^2-x-1}{2\sqrt{1-x^2}}$

A'=$0$ When $x$ = - $\dfrac{1}{2}$

$y$=$\dfrac{\sqrt{3}}{2}$

Plugging everything back into the original area equation I get $\dfrac{ \sqrt{3}}{8}$

Answer 1

Almost: Since the area is

$$ A = \frac{(1+x)\sqrt{1-x^2}}{2} $$

we want to find the zero of the derivative

$$ \frac{dA}{dx} = \frac{1-x-2x^2}{2\sqrt{1-x^2}} $$

Since $-1 < x < 1$ (do you see why?), we can just focus on the numerator, and then

$$ 2x^2+x-1 = 0 $$

or $x = -1$ or $x = 1/2$. The first solution is discarded, and we use the second solution: $x = 1/2$, so $y = \sqrt{3}/2$, and $A = (1+x)y/2 = 3\sqrt{3}/8 \doteq 0.64952$.

You just lost a minus sign, is all, I think. But it is useful to evaluate your answer to see if it makes sense. $\sqrt{3}/8 \doteq 0.21651$ is too small for something that must be larger than half the unit square.

Answer 2

Another approach is to use $\angle CBD=\phi\in(0,\frac\pi2)$:

\begin{align} |BC|&=|AB|\cos\phi=2\cos\phi \\ |BD|&=|BC|\cos\phi=2\cos^2\phi \\ S_{\triangle BCD}&= \tfrac12|BD|\cdot|BC|\sin\phi \\ &= 2\sin\phi\cos^3\phi \\ S'(\phi) &= 2\cos^2\phi(\cos^2\phi-3\sin^2\phi) \\ &= 2\cos^2\phi(1-4\sin^2\phi) \end{align} The only suitable solution to $S'(\phi)=0$ is $\phi=\arcsin\tfrac12=\tfrac\pi6$, hence \begin{align} S_{\max} &= 2\cdot\tfrac12\cdot\left(\tfrac{\sqrt3}{2}\right)^3 =\tfrac{3\sqrt3}{8}. \end{align}