smc

Complete the square to get $$(x-7{)}^2+(y-3{)}^2=64.$$ Applying Cauchy-Schwarz directly, $$64\cdot 25=(3^2+4^2)((x-7{)}^2+(y-3{)}^2)\ge (3(x-7)+4(y-3){)}^2.$$ $$40\ge 3x+4y-33$$ $$3x+4y\le 73.$$ Thus our answer is $\boxed{73}$.