Lagrange Multipliers — Question 3

Question 3 – Solution

Strategy. Minimize squared distance, which has the same minimizers as distance but simpler derivatives.

Step 1: Set up Let $$f(x,y)=x^2+(y-2)^2,\qquad g(x,y)=y-x^2.$$ The system $\nabla f=\lambda\nabla g$ is $$2x=-2\lambda x,\qquad 2(y-2)=\lambda,\qquad y=x^2.$$

Step 2: Solve branches If $x=0$, then $y=0$, producing the candidate $(0,0)$ with squared distance $4$.

If $x\ne 0$, the first equation gives $\lambda=-1$. The second gives $2(y-2)=-1$, so $y=3/2$. Hence $$x=\pm\sqrt{\frac 32}.$$ At either point the squared distance is $$\frac 32+\left(-\frac 12\right)^2=\boxed{\frac 74}.$$

Step 3: Conclusion Comparing candidates gives the two closest points $$\boxed{\left(\pm\sqrt{\frac 32},\frac 32\right)},$$ each at distance $\boxed{\sqrt 7/2}$ from $A$. Along the parabola, $f(x,x^2)=x^2+(x^2-2)^2\to\infty$ as $|x|\to\infty$. Therefore no minimizing sequence can escape to infinity, and the smaller candidate value is the absolute minimum.