Lagrange Multipliers — Question 6

Question 6 – Solution

Strategy. Inspect the feasible set before applying the multiplier equation; the constraint has a singular gradient at its only point.

Step 1: Direct analysis The equation $x^2+y^2=0$ over the reals forces $$x=0,\qquad y=0.$$ Thus the feasible set is the singleton $\{(0,0)\}$. Its only function value is $0$, so $$\boxed{f_{\min}=f_{\max}=0\text{ at }(0,0)}.$$

Step 2: Multiplier failure $$\nabla f(0,0)=\left\langle 1,0\right\rangle, \qquad \nabla g(0,0)=\left\langle 0,0\right\rangle.$$ The equation $$\left\langle 1,0\right\rangle=\lambda\left\langle 0,0\right\rangle$$ has no solution for any real $\lambda$.

Step 3: Interpretation The standard Lagrange multiplier theorem assumes $\nabla g\ne\mathbf 0$ at the constrained extremum. That regularity condition fails here, so the theorem makes no promise. The example shows that singular feasible points must be checked separately; it does not invalidate the multiplier method at regular constraint points.