Given that $x^2 + y^2 = 14x + 6y + 6$, what is the largest possible value that $3x + 4y$ can have? 
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$
Complete the square to get 
\[(x-7)^2 + (y-3)^2 = 64.\]
Applying Cauchy-Schwarz directly,
\[64\cdot25=(3^2+4^2)((x-7)^2 + (y-3)^2) \ge (3(x-7)+4(y-3))^2.\]
\[40 \ge 3x+4y-33\]
\[3x+4y \le 73.\]
Thus our answer is $\boxed{(B)}$.