Calculus: Maximum and minimum

The diagram below show three right-angle triangles, where $\displaystyle BC=14,GF=10,DE=7$ and $\displaystyle \angle BCA=\angle BDE=\angle FGD=\theta$ . Find the maximum possible value of $\displaystyle AB+BD+DF$.

Solution

$\displaystyle BC=14,GF=10,DE=7$

$\displaystyle \angle BCA=\angle BDE=\angle FGD=\theta$

$\displaystyle \text{Let}\ \ \ y=AB+BD+DF$

$\displaystyle \ \ \ \ \ \ \ \ \ \ =14\sin \theta +7\cos \theta +10\sin \theta$

$\displaystyle \ \ \ \ \ \ \ \ \ \ =24\sin \theta +7\cos \theta$

$\displaystyle \ \ \ \ \ \frac{{dy}}{{d\theta }}=24\cos \theta -7\sin \theta$

$\displaystyle \text{For}\ \text{stationary}\ \text{value}\ ,\ \frac{{dy}}{{d\theta }}=0$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ 24\cos \theta -7\sin \theta =0$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 24\cos \theta =7\sin \theta$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \tan \theta =\frac{{24}}{7}$

$\displaystyle \ \ \ \ \ \frac{{{{d}^{2}}y}}{{d{{\theta }^{2}}}}=-24\sin \theta -7\cos \theta$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =-\cos \theta \ (24\tan \theta -7)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =-\cos \theta \ (\frac{{576}}{7}-7)$

$\displaystyle \text{Since }\ \theta \ \ \text{is}\ \text{acute}\ ,\ \frac{{{{d}^{2}}y}}{{d{{\theta }^{2}}}}<0\ (\text{maximum})$

$\displaystyle \text{The}\ \text{maximum}\ \text{possible}\ \text{value}\ \text{of}\ AB+BD+DF$

$\displaystyle =24\sin \theta +7\cos \theta$

$\displaystyle =24(\frac{{24}}{{25}})+7(\frac{7}{{25}})$

$\displaystyle =\frac{{576}}{{25}}+\frac{{49}}{{25}}$

$\displaystyle =\frac{{625}}{{25}}=25$