Function Optimization with Non-linear Constraint, Lagrange Multipliers Fails I am trying to maximize the function $A(x,y)=\frac{1}{2}(x(12-x)+y(13-y))$ subject to the constraint $x^2+(12-x)^2-y^2-(13-y)^2=0$.
My attempt:
$\begin{align*} \nabla A=\frac{1}{2}\langle 12-2x,\,13-2y\rangle &= \lambda\langle4x-24,\, -4y+26\rangle\\ \implies&\begin{cases} -x+6=\lambda(4x-24)\\-y+\frac{13}{2}=\lambda(-4y+26)\\x^2+(12-x)^2-y^2-(13-y)^2=0\end{cases}\end{align*}$
But clearly there is no solution due to the first two equations.
Using Wolfram Alpha, however, yields a maximum at $\displaystyle \left(\frac{17}{2},\,\frac{13}{2}\right)$ being $A=36$ and shows a nice little graph.

Let $x(12-x)=a$ and $y(13-y)=b$.
Thus, the condition gives $b=a+12.5$.
Also, we have
$$a=x(12-x)\leq\left(\frac{x+12-x}{2}\right)=36$$ and
$$b=y(13-y)\leq\left(\frac{y+13-y}{2}\right)=42.25,$$
which gives
$$a=b-12.5\leq42.25-12.5=29.75.$$
Id est, $$A(x,y)=\frac{1}{2}(a+b)=a+6.25\leq29.75+6.25=36.$$
The equality occurs for $b=42.25$ or $y=6.5$, which says that we got a maximal value.
