jmc

Moving all the terms to the left, we have the equation $x^2-14x+y^2-48y=0$. Completing the square on the quadratic in $x$, we add $(14/2{)}^2=49$ to both sides. Completing the square on the quadratic in $y$, we add $(48/2{)}^2=576$ to both sides. We have the equation $$(x^2-14x+49)+(y^2-48y+576)=625\Rightarrow (x-7{)}^2+(y-24{)}^2=625$$ Rearranging, we have $(y-24{)}^2=625-(x-7{)}^2$. Taking the square root and solving for $y$, we get $y=\pm \sqrt{625-(x-7{)}^2}+24$. Since $\sqrt{625-(x-7{)}^2}$ is always nonnegative, the maximum value of $y$ is achieved when we use a positive sign in front of the square root. Now, we want the largest possible value of the square root. In other words, we want to maximize $625-(x-7{)}^2$. Since $(x-7{)}^2$ is always nonnegative, $625-(x-7{)}^2$ is maximized when $(x-7{)}^2=0$ or when $x=7$. At this point, $625-(x-7{)}^2=625$ and $y=\sqrt{625}+24=49$. Thus, the maximum $y$ value is $\boxed{49}$.

--OR--

Similar to the solution above, we can complete the square to get the equation $(x-7{)}^2+(y-24{)}^2=625$. This equation describes a circle with center at $(7,24)$ and radius $\sqrt{625}=25$. The maximum value of $y$ is achieved at the point on the top of the circle, which is located at $(7,24+25)=(7,49)$. Thus, the maximum value of $y$ is $\boxed{49}$.