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The objective is continuous on the compact disk, so a global maximum exists. Its gradient is $(e^x,2e^{2y})$, which is never zero, so a maximum cannot lie in the interior. It must have $x^2+y^2=1$.

Both coordinates of a maximizing point are positive. Reflecting a negative coordinate to its positive counterpart strictly increases the objective without changing the constraint; at an endpoint with one coordinate zero, a small move along the circle into the positive quadrant also increases the objective to first order. Thus it suffices to consider $x>0,y>0$.

The boundary Lagrange multiplier equations are $e^x=2\lambda x$ and $2e^{2y}=2\lambda y$. Eliminate $\lambda$ to obtain the two-dimensional nonlinear system of equationsKeeping this quadrant restriction excludes the boundary minimum that an unconstrained elimination could also produce. Under the stated uniqueness assumption for the admissible solution, the global maximum must be that solution. Approximate it with multivariable Newton iteration, solving $J_F(z_n){\Delta }_n=-F(z_n)$ and setting $z_{n+1}=z_n+{\Delta }_n$, with a suitable safeguard if needed. For example its explicit Jacobian matrix isNo numerical solution of the system is required.

A system of equations requires one choice of unknowns to satisfy several equations at once. Its solution set is the intersection of the solution sets of the individual equations. For example, $x^2+y^2=1$ and $y=(x-1{)}^2$ ask for intersections of a unit circle and a parabola. A linear system is the special case in which every equation is linear in the unknowns.