Help Please

As follows:

Here's a non calculus approach.

Begin by noticing that

$7^{2}+24^{2}=49+576=625=25^{2},$

so 7, 24, 25 can be used to as the sides of a right-angled triangle.

Let the angle opposite the 7 be $\phi,$

then

$\sin\phi=7/25 \text{ and } \cos\phi=24/25.$

$\displaystyle 7\cos\theta+24\sin\theta+7\\ = 25(\frac{7}{25}\cos\theta+\frac{24}{25}\sin\theta)+7\\ =25(\sin\phi\cos\theta+\cos\phi\sin\theta)+7 \\ =25\sin(\theta+\phi)+7.$

Maximum value will be 25 + 7 = 32 occuring when

 $\sin(\theta+\phi)=1, \\ \theta=\pi/2-\tan^{-1}(7/24),\\ 0\leq\theta\leq\pi/2.$