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Mathematics 15 Online
OpenStudy (anonymous):

find the maximum value of

OpenStudy (anonymous):

\[\sin 2 \alpha+\sin 2 \beta+\sin 2\gamma \] where \[\alpha,\beta,\gamma \ge 0,\text{and } \alpha+\beta+\gamma=180\]

OpenStudy (amistre64):

sounds like a good candidate for a lagrange multiplier run ... maybe

OpenStudy (amistre64):

and we say that a+b+g = pi to convert degrees to radians, not too sure if that matters tho

OpenStudy (amistre64):

\[f(x,y,z)\sin 2x+\sin 2y+\sin 2z~:~g(x,y,z)=x+y+z-k\] \[f_x2cos(2x)~:~\lambda g_x=\lambda~:~x=\frac12cos^{-1}\left(\frac{\lambda}{2}\right)\] \[f_x2cos(2y)~:~\lambda g_y=\lambda\] \[f_x2cos(2z)~:~\lambda g_z=\lambda\] looks like x=y=z if ive done it correctly

OpenStudy (amistre64):

\[\frac32cos^{-1}\left(\frac{\lambda}{2}\right)=\pi(or~180?)\] \[\lambda=2cos\left(\frac23\pi(or~180?)\right)\]

OpenStudy (amistre64):

\[\lambda=2cos\left(\frac23\pi\right)\] \[\frac{\lambda}{2}=cos\left(\frac23\pi\right)\] \[cos^{-1}\frac{\lambda}{2}=\frac23\pi\] \[\frac12cos^{-1}\frac{\lambda}{2}=\frac\pi 3\]

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