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$$\begin{array}{ll} \underset{a,b,c,d,e}{\text{maximize}} & \sin (a) + \sin (b) + \sin (c) + \sin (d) + \sin (e) \\ \text{subject to} & a + b + c + d + e = 2\pi \end{array}$$

How do I prove that the objective is maximized only when $a = b = c = d = e = \frac{2\pi}{5}$?

Rodrigo de Azevedo
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Draculin
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  • Seems like a job for induction, although I could be mistaken. Suppose that $\sin(r) > 0$ and you wanted to choose a value of $s$ so as to maximize the sum $$\sin(r+s) + \sin(r-s) = 2\sin(r)\cos(s).$$ What value of $s$ would you choose? – user2661923 Apr 22 '22 at 10:41
  • If you know Lagrange multipliers, the answer is immediate. – Claude Leibovici Apr 22 '22 at 10:46

1 Answers1

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Using the method of Lagrange multipliers you immediately get to this. Let me first define $g=a+b+c+d+e-2\pi=0$ then the lagrange function is $$L(a,b,c,d,\lambda)=E-\lambda g=\sin(a)+\sin(b)+\sin(c)+\sin(d)+\sin(e)-\lambda(a+b+c+d+e-2\pi)$$ Now for lagrange you need to compute all partial derivatives. $$\frac{\partial L}{\partial x}=\cos(x)-\lambda$$ for $x\in \{a,b,c,d\}$ and $$\frac{\partial L}{\partial \lambda}=a+b+c+d+e=2\pi$$ Then you only need to solve the system$$\frac{\partial L}{\partial x}=\cos(x)-\lambda=0\Leftrightarrow \cos(x)=\lambda$$ for $x\in \{a,b,c,d\}$ and you immediately get that $$a=b=c=d=e=\frac{2\pi}{5}$$

I hope this helps.

user123234
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