Lagrange Multipliers — Question 8

Question 8 – Solution

Strategy. Parallel gradients at the specified regular point determine the parameter; a one-variable restriction then establishes the classification.

Step 1: Recover $a$ For $g=x+y$, $$\nabla f_a(1,2)=\left\langle 2,4a\right\rangle, \qquad \nabla g=\left\langle 1,1\right\rangle.$$ The equation $\left\langle 2,4a\right\rangle=\lambda\left\langle 1,1\right\rangle$ gives $\lambda=2$ and $4a=2$. Hence $$\boxed{a=\frac 12}.$$

Step 2: Restrict to the line Set $y=3-x$. Then $$f_{1/2}(x,3-x)=x^2+\frac 12(3-x)^2.$$ Completing the square, $$x^2+\frac 12(3-x)^2 =\frac 32(x-1)^2+3.$$ Therefore the constrained value is minimized uniquely at $x=1$, $y=2$.

Step 3: Result and verification $$\boxed{P=(1,2)\text{ is a strict constrained minimum}}, \qquad \boxed{f_{1/2}(1,2)=3}.$$ The restricted quadratic tends to infinity as $|x|\to\infty$, so the minimum is also absolute along the entire constraint line; no constrained maximum exists.