Question Number 106941 by mr W last updated on 08/Aug/20
$${Find}\:{the}\:{maximum}\:{value}\:{of}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{sin}^{\mathrm{5}} \:\theta_{{i}} \\ $$$${with}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{sin}\:\theta_{{i}} =\mathrm{0}. \\ $$
Answered by mr W last updated on 08/Aug/20
![let x_i =sin θ_i −1≤x_i ≤1 let x_1 =x_2 =...=x_m =1, ⇒x_(m+1) =x_(m+2) =...=x_n =−(m/(n−m)) Sw=Σ_(i=1) ^n sin^5 θ_i =m×1^5 +(n−m)×(−(m/(n−m)))^5 S=m−(m^5 /((n−m)^4 ))=m[1−((m/(n−m)))^4 ] (dS/dm)=1−((5m^4 )/((n−m)^4 ))−((4m^5 )/((n−m)^5 ))=0 n^3 −5n^2 m+10nm^2 −10m^3 =0 10((m/n))^3 −10((m/n))^2 +5((m/n))−1=0 ⇒(m/n)≈0.3773 ⇒m≈0.3773n S_(max) ≈0.3773n[1−(((0.3773)/(1−0.3773)))^4 ]≈0.3265n](Q107086.png)
$${let}\:{x}_{{i}} =\mathrm{sin}\:\theta_{{i}} \\ $$$$−\mathrm{1}\leqslant{x}_{{i}} \leqslant\mathrm{1} \\ $$$${let}\:{x}_{\mathrm{1}} ={x}_{\mathrm{2}} =...={x}_{{m}} =\mathrm{1}, \\ $$$$\Rightarrow{x}_{{m}+\mathrm{1}} ={x}_{{m}+\mathrm{2}} =...={x}_{{n}} =−\frac{{m}}{{n}−{m}} \\ $$$${Sw}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{sin}^{\mathrm{5}} \:\theta_{{i}} ={m}×\mathrm{1}^{\mathrm{5}} +\left({n}−{m}\right)×\left(−\frac{{m}}{{n}−{m}}\right)^{\mathrm{5}} \\ $$$${S}={m}−\frac{{m}^{\mathrm{5}} }{\left({n}−{m}\right)^{\mathrm{4}} }={m}\left[\mathrm{1}−\left(\frac{{m}}{{n}−{m}}\right)^{\mathrm{4}} \right] \\ $$$$\frac{{dS}}{{dm}}=\mathrm{1}−\frac{\mathrm{5}{m}^{\mathrm{4}} }{\left({n}−{m}\right)^{\mathrm{4}} }−\frac{\mathrm{4}{m}^{\mathrm{5}} }{\left({n}−{m}\right)^{\mathrm{5}} }=\mathrm{0} \\ $$$${n}^{\mathrm{3}} −\mathrm{5}{n}^{\mathrm{2}} {m}+\mathrm{10}{nm}^{\mathrm{2}} −\mathrm{10}{m}^{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{10}\left(\frac{{m}}{{n}}\right)^{\mathrm{3}} −\mathrm{10}\left(\frac{{m}}{{n}}\right)^{\mathrm{2}} +\mathrm{5}\left(\frac{{m}}{{n}}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{{m}}{{n}}\approx\mathrm{0}.\mathrm{3773} \\ $$$$\Rightarrow{m}\approx\mathrm{0}.\mathrm{3773}{n} \\ $$$${S}_{{max}} \approx\mathrm{0}.\mathrm{3773}{n}\left[\mathrm{1}−\left(\frac{\mathrm{0}.\mathrm{3773}}{\mathrm{1}−\mathrm{0}.\mathrm{3773}}\right)^{\mathrm{4}} \right]\approx\mathrm{0}.\mathrm{3265}{n} \\ $$