Maximum Value

Solution 1

Using Cauchy-Schwarz ,

$(3a+4b)^2\leq(3^2+4^2)(a^2+b^2)=225$

Since we find the maximum value we take the positive root $3a+4b\leq15$

$\therefore 3a+4b+5\leq \boxed{20}$

Solution 2

Substitute $a=\sqrt{9-b^2}$ into $3a+4b+5$ ,

$3a+4b+5$

$=3\sqrt{9-b^2} + 4b + 5$

Taking the derivative yields

$\frac{d}{db}(3\sqrt{9-b^2}+4b+5)$

$=\frac{3b}{\sqrt{9-b^2}} + 4 = 0$

Solving for $b$ , $b=\frac{12}{5}$ and it is a maximum value after second derivative test.

Substituting yields $3a+4b+5=\boxed{20}$