Suppose one wants to maximise the function
$$U_1 (x) = (x_1 + 1) x_2$$
subject to the constraints
$$U_2 (x) = 36 − (x_1 − 4)^2 − (x_2 − 6)^2 = c$$
$$0 \leq x_1 \leq 3$$
$$0 \leq x_2 \leq 5$$
where $U_1$ and $U_2$ are utility functions of two persons. I understand how to solve a simple Lagrange multiplier situation, but in this situation, I don't understand the best technique to use the information constraining the values that $x_1$ and $x_2$ can take. I understand that the constraint here plots a circle of radius $(36-c)^\frac{1}{2}$, and that the inequalities limiting the value of $x_1$ and $x_2$ describe where that circle can be placed.
Trying to solve the equation without the input of inequalities is tedious, and leads to large and difficult equations once $\lambda$ is removed.
Any guidance on how to proceed and solve this equation would be much appreciated.
The main constraint defines a circle $\gamma$ with center $(4,6)$ and radius $r:=\sqrt{36-c}$, whereas the inequality constraints restrict the feasible set to the rectangle $R:=[0,3]\times[0,5]$. If $r<\sqrt{2}$ or $r>\sqrt{52}$ there are no feasible points. If $\sqrt{2}\leq r\leq\sqrt{52}$ the set $\Phi:=\gamma\cap R$ of feasible points is an arc $\gamma'$ of $\gamma$, hence essentially an interval. The endpoints $A$, $B$ of $\gamma'$ are obtained by intersecting $\gamma$ with $\partial R$.
Now the function $U_1$ takes its maximum on $\Phi$ either in a boundary point of $\gamma'$, or in an interior point $P$ of $\gamma'$. In the latter case the point $P$ is brought to the fore by applying Lagrange's method. The method will produce a certain number of points $P_k\in\gamma$, some of them not in $\gamma'$. Throw the latter away. The maximum value of $f:=U_1$ under the given conditions is then the maximum value among $f(A)$, $f(B)$, and the $f(P_k)$ for the $P_k\in\gamma'$.
A lot of work indeed, in particular if $c$ is not specified in advance.