$\frac{x^2}{4}+\frac{y^2}{49}=1$

$$ Note that, if we plot $\dfrac{x^2}{4}+\dfrac{y^2}{49}=1$ , it will be an ellipse and if we plot $x+y=\sqrt{k}$ , it will be a straight line. The largest possible value of $\sqrt{k}$ for which the line meets the ellipse at a $\sqrt{k}$ for which the line is tangent to the ellipse. So the maximum occurs when the straight line and the ellipse meet at a single point. From $x+y=\sqrt{k}$ we have $y=-x+\sqrt{k}$ and substitute this to ellipse equation, yield $$ $\begin{aligned} \frac{x^2}{4}+\frac{(-x+\sqrt{k})^2}{49}&=1\\ \frac{x^2}{4}+\frac{x^2-(2\sqrt{k})x+k}{49}&=1\\ 49x^2+4x^2-(8\sqrt{k})x+4k&=4\cdot49\\ 53x^2-(8\sqrt{k})x+4k-196&=0. \end{aligned}$ $$ The straight line and the ellipse will meet at a single point if the discriminant of the last quadratic equation is equal to zero. Therefore $$ $\begin{aligned} (-8\sqrt{k})^2-4\cdot53(4k-196)&=0\\ 64k-848k+41552&=0\\ k&=\frac{41552}{784}\\ &=\boxed{53} \end{aligned}$ $$ Another approach, we can solve this problem by using Lagrange multiplier .