Lagrange Multipliers — Question 7

Question 7 – Solution

Strategy. Eliminate the multiplier by combining the two component equations; the resulting factorization identifies the principal directions.

Step 1: Multiplier equations $$\left\langle 2x,2y\right\rangle=\lambda\left\langle 2x+y,x+2y\right\rangle, \qquad x^2+xy+y^2=3.$$ Multiply the first component equation by $y$, the second by $x$, and subtract. The left sides cancel, leaving $$\lambda(y^2-x^2)=0.$$ The constraint rules out $(0,0)$, and $\lambda=0$ would force $x=y=0$. Hence $$y=\pm x.$$

Step 2: Candidate values If $y=x$, the constraint is $3x^2=3$, so $(x,y)=(1,1)$ or $(-1,-1)$ and $f=2$.

If $y=-x$, the constraint is $x^2=3$, giving $(\sqrt 3,-\sqrt 3)$ and $(-\sqrt 3,\sqrt 3)$ with $f=6$. Therefore $$\boxed{f_{\min}=2\text{ at }(1,1),(-1,-1)},$$ $$\boxed{f_{\max}=6\text{ at }(\sqrt 3,-\sqrt 3),(-\sqrt 3,\sqrt 3)}.$$

Step 3: Verification The constraint ellipse is compact. Along its $y=x$ principal direction its radius squared is $2$, while along $y=-x$ its radius squared is $6$. These are exactly the values of $x^2+y^2$, confirming the classifications and completeness.