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Question Number 106941 by mr W last updated on 08/Aug/20

$$Findthemaximumvalueof\underset{i=1}{\sum ^n}{\sin }^5{\theta }_i$$$$with\underset{i=1}{\sum ^n}\sin {\theta }_i=0.$$

Answered by mr W last updated on 08/Aug/20

$$let{x}_i=\sin {\theta }_i$$$$-1⩽x_i⩽1$$$$let{x}_1=x_2=...=x_m=1,$$$$\Rightarrow x_{m+1}=x_{m+2}=...=x_n=-\frac{m}{n-m}$$$$Sw=\underset{i=1}{\sum ^n}{\sin }^5{\theta }_i=m\times 1^5+(n-m)\times {(-\frac{m}{n-m})}^5$$$$S=m-\frac{m^5}{{(n-m)}^4}=m[1-{\left(\frac{m}{n-m}\right)}^4]$$$$\frac{dS}{dm}=1-\frac{5m^4}{{(n-m)}^4}-\frac{4m^5}{{(n-m)}^5}=0$$$$n^3-5n^{2}m+10{nm}^2-10m^3=0$$$$10{\left(\frac{m}{n}\right)}^3-10{\left(\frac{m}{n}\right)}^2+5\left(\frac{m}{n}\right)-1=0$$$$\Rightarrow \frac{m}{n}\approx 0.3773$$$$\Rightarrow m\approx 0.3773n$$$$S_{max}\approx 0.3773n[1-{\left(\frac{0.3773}{1-0.3773}\right)}^4]\approx 0.3265n$$

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