calculusderivativesgeometryoptimization
The sum of the lengths of the hypotenuse and another side of a right angled triangle is given.The area of the triangle will be maximum if the angle between them is:
$(A)\frac{\pi}{6}\hspace{1cm}(B)\frac{\pi}{4}\hspace{1cm}(C)\frac{\pi}{3}\hspace{1cm}(D)\frac{5\pi}{12}$
Let $a,b$ are sides of a right triangle other than hypotenuse.Then given that $\sqrt{a^2+b^2}+a$=constant=$k$
Area of triangle=$\frac{1}{2}ab=\frac{1}{2}a\sqrt{(k-a)^2-a^2}$
and then?
The answer is $84$. It comes from the right triangle $(12,35,37)$.
You want the maximum of $12+a+b$ where $a$ and $b$ are such that $12^2+a^2=b^2$. This is equivalent to $(b-a)(b+a)=144$. Making the list of all possible ways of expressing $144$ as the product of two numbers both of which are even, you will find that the list is rather short. My answer comes from the fact that $144=2\times72$. For this decomposition, you get $a=35$ and $b=37$.
It can be easily proved by geometry
Consider a right $\Delta ABC$ having hypotenuse $AC$ of constant length which is inscribed in a semi-circle with center $O$ & radius $OA=OE=OC(=AC/2)$ (as shown in figure below).
The right angled vertex $B$ can lie anywhere on the semi-circular arc AC such that $$BD\le EO$$ $$BD\le \color{blue}{\frac{AC}{2}}$$
Best Answer
The area of the triangle can be expressed as $$\frac 12ab=\frac 12a\sqrt{(k-a)^2-a^2}=\frac{a}{2}\sqrt{k^2-2ak}=\frac{\sqrt k}{2}\sqrt{a^2k-2a^3}$$
Here, let $f(a)=a^2k-2a^3$. Then, we have $$f'(a)=2ak-6a^2=2a(k-3a).$$ So, since we know that the maximum of $f(a)$ is $f(k/3)$, it follows that the maximum of the area of the trangle is attained when $a=k/3$.
Then, the angle $\theta$ satisfies $$\cos\theta=\frac{a}{\sqrt{a^2+b^2}}=\frac{k/3}{k-(k/3)}=\frac 12\Rightarrow \theta=\frac{\pi}{3}.$$