(b) Hence or otherwise find the maximum value of $ \displaystyle f(x)$ where $ \displaystyle f(x)=\frac{{30}}{{5\sin x+12\cos x+17}}$ .
State the values of x, in the range 0° < x < 360°, at which they occur.
Solution
Let $ \displaystyle 5\sin x+12\cos x=R\sin (x+\theta ),$ where $ \displaystyle R\cos \theta =5$ and $ \displaystyle R\sin \theta =12.$
$ \displaystyle \therefore {{(R\cos \theta )}^{2}}+{{(R\sin \theta )}^{2}}={{5}^{2}}+{{12}^{2}}$
$ \displaystyle \therefore {{R}^{2}}{{\cos }^{2}}\theta +{{R}^{2}}{{\sin }^{2}}\theta =169$
$ \displaystyle \therefore {{R}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )=169$
$ \displaystyle \therefore {{R}^{2}}(1)=169$
$ \displaystyle \therefore R=\sqrt{{169}}=13$
And $ \displaystyle \frac{{R\sin \theta }}{{R\cos \theta }}=\frac{{12}}{5}\Rightarrow \tan \theta =2.4\Rightarrow \theta =67{}^\circ 2{3}'\text{ }$
$ \displaystyle \therefore 5\sin x+12\cos x=13\sin (x+67{}^\circ 2{3}')$
$ \displaystyle f(x)=\frac{{30}}{{5\sin x+12\cos x+17}}$ (given)
$ \displaystyle \ \ \ \ \ \ \ =\frac{{30}}{{13\sin (x+67{}^\circ 2{3}')+17}}$
$ \displaystyle \ \ \ \ \ \ \ =\frac{{30}}{{g(x)+17}}$ where $ \displaystyle g(x)=13\sin (x+67{}^\circ 2{3}')$
Therefore $ \displaystyle f(x)$ is maximum when $ \displaystyle g(x)$ is minimum and vice versa.
Since $ \displaystyle -1\le \sin (x+67{}^\circ 2{3}')\le 1$,
$ \displaystyle -13\le 13\sin (x+67{}^\circ 2{3}')\le 13$
$ \displaystyle -13\le g(x)\le 13$
Hence the minimum value of $ \displaystyle g(x)=-13$
$ \displaystyle \therefore 13\sin (x+67{}^\circ 2{3}')=-13$
$ \displaystyle \ \ \ \sin (x+67{}^\circ 2{3}')=-1$
$ \displaystyle \ \ \ x+67{}^\circ 2{3}'=270{}^\circ $
$ \displaystyle \therefore x=202{}^\circ 3{7}'$
$ \displaystyle \therefore x=202{}^\circ 3{7}'$
Therefore the maximum value of $ \displaystyle f(x)$ is $ \displaystyle \frac{{30}}{{-13+17}}=\frac{{15}}{2}$ when $ \displaystyle x=202{}^\circ 3{7}'.$
The maximum value of $ \displaystyle g(x)=13.$
$ \displaystyle \therefore 13\sin (x+67{}^\circ 2{3}')=13$
$ \displaystyle \ \ \ \sin (x+67{}^\circ 2{3}')=1$
$ \displaystyle \ \ \ x+67{}^\circ 2{3}'=90{}^\circ $
$ \displaystyle \therefore x=22{}^\circ 3{7}'$
Therefore the minimum value of $ \displaystyle f(x)$ is $ \displaystyle \frac{{30}}{{13+17}}=1$ when $ \displaystyle x=22{}^\circ 3{7}'.$
Illustration
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