Exercice XV num 9 - Calculus Made Easy

Question

Divide π into 3 parts such that the product of their sines may be a maximum or minimum. What is the most intuitive way to solve it?

Answer 1

Let $f:[0,\pi]\times[0,\pi]\to\mathbb{R}$, $f(x,y)=\sin x \sin y\sin(\pi-x-y)$. You have $\frac{\partial f}{\partial x} =\cos x\sin y\sin(\pi-x-y)-\sin x \sin y \cos(\pi-x-y)$ and $\frac{\partial f}{\partial y} =\sin x\cos y\sin(\pi-x-y)-\sin x \sin y \cos(\pi-x-y)$. Set both equal to zero and you find, after dividing by $\cos x $ and $ \cos(\pi-x-y)$ the first and by $\cos y$ and $\cos(\pi-x-y)$ the second, $\tan x=\tan y=\tan(\pi-x-y)$. Hence $x=y=\pi/3$.

Answer 2

I think it works best with Lagrange multipliers. Let $f(A,B,C) =\sin(A)\sin(B)\sin(C)$ and let $g(A,B,C) = A+B+C = \pi$ be the constraint. The system of equations $\nabla f =\lambda \nabla g$ is:

$$\cos(A)\sin(B)\sin(C) = \lambda$$

$$\sin(A)\cos(B)\sin(C) = \lambda$$

$$\sin(A)\sin(B)\cos(C) =\lambda$$

Subtract the first two equations and cancel $\sin(C)$ to get

$$\cos(A)\sin(B) - \sin(A)\cos(B) = 0$$

or $$\sin(A-B) = 0.$$

Assuming $0<A,B<\pi$ we must have $A=B$.

Similarly we have $B=C$. So the solution is $3A=\pi$ or $A=B=C=\pi/3$.