jmc

In the equation $$3x^2-18x+4y^2-32y+91=300,$$we can complete the square in $x$ and $y$ to get $$3(x-3{)}^2+4(y-4{)}^2=300.$$We want to find the maximum value of $$x^2+y^2+2xy-14x-14y+49=(x+y{)}^2-14(x+y)+49=(x+y-7{)}^2.$$By Cauchy-Schwarz, $$\left(\frac{1}{3}+\frac{1}{4}\right)[3(x-3{)}^2+4(y-4{)}^2]\ge ((x-3)+(y-4){)}^2=(x+y-7{)}^2,$$so $$(x+y-7{)}^2\le \frac{7}{12}\cdot 300=175.$$Equality occurs when $3(x-3)=4(y-4)$ and $3(x-3{)}^2+4(y-4{)}^2=300.$ We can solve to get $x=\frac{1}{7}(21\pm 20\sqrt{7})$ and $y=\frac{1}{7}(28\pm 15\sqrt{7}),$ so the maximum value is $\boxed{175}.$