Maximum

We know that $AB = 14\sin \theta$

$BD = 7 \cos \theta$

$DF = 10 \sin \theta$

Adding them, we get the desired expression as $24 \sin \theta + 7 \cos \theta$ We need to find the maximum of this.

By Cauchy - Swartz, we know that

$(24^2 + 7^2)(\sin^2 \theta + \cos^2 \theta) \geq (24 \sin \theta + 7 \cos \theta)^2$

$\implies 25 \geq 24 \sin \theta + 7 \cos \theta$

Equality holds iff $\theta = \tan^{-1}{\frac{24}{7}}$

Therefore $\boxed{25}$ is the maximum.

$\begin{aligned} AB+BD+DF & = BC\sin \theta + DE\cos \theta + GF\sin \theta \\ & = 14 \sin \theta + 7 \cos \theta + 10 \sin \theta \\ & = 24 \sin \theta + 7 \cos \theta \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Note that }7^2+24^2=25^2} \\ & = 25 \left( \frac{24}{25} \sin \theta + \frac{7}{25} \cos \theta \right) \quad \quad \small \color{#3D99F6}{\text{Let } \tan \phi = \frac{7}{24}} \\ & = 25 \left(\sin \theta \cos \phi + \sin \phi \cos \theta \right) \\ & = 25 \sin (\theta + \phi) \end{aligned}$

Since maximum of $\sin (\theta + \phi) = 1$ , therefore maximum of $AB+BD+DF = \boxed{25}$ .

Assume the random value of AB+BD+DF is n and the answer is m. Notice that with the given data:

$n=AB+BD+DF=14\sin\theta+7\cos\theta+10\sin\theta=24\sin\theta+7\cos\theta=(24\tan\theta+7)\cos\theta$

$\frac{dn}{d\theta}=\frac{d}{d\theta}(24\sin\theta+7\cos\theta)=24\cos\theta-7\sin\theta$

When this first derivative equals zero, since the value of n is upper and lower bounded, then there is a value of $\theta$ that maximize the value of n. We get $24\cos\theta=7\sin\theta$ , which means $\tan\theta=\frac{24}{7}$ and this, with $\cos^2\theta+\sin^2\theta=1$ , results $\cos\theta=\pm\frac{7}{\sqrt{24^2+7^2}}=\pm\frac{7}{25}$ . We take the positive value because it makes n positive and thus maximum. Hence:

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