Lagrange Multipliers — Question 2

Question 2 – Solution

Strategy. Solve the component equations without dividing until zero-coordinate cases have been checked.

Step 1: Multiplier equations With $g=x^2+4y^2$, $$\left\langle y,x\right\rangle=\lambda\left\langle 2x,8y\right\rangle, \qquad x^2+4y^2=4.$$ If $x=0$, the first multiplier equation forces $y=0$, violating the constraint; similarly $y=0$ is impossible. Thus $xy\ne 0$, and $$y=2\lambda x,\qquad x=8\lambda y.$$ Consequently $16\lambda^2=1$ and $y=\pm x/2$. The constraint then gives $2x^2=4$, so $x=\pm\sqrt 2$.

Step 2: Candidates and values Equal signs for $x$ and $y$ give $xy=1$; opposite signs give $xy=-1$. Therefore $$\boxed{f_{\max}=1\text{ at }(\sqrt 2,\sqrt 2/2), (-\sqrt 2,-\sqrt 2/2)},$$ $$\boxed{f_{\min}=-1\text{ at }(\sqrt 2,-\sqrt 2/2), (-\sqrt 2,\sqrt 2/2)}.$$

Step 3: Verification Let $u=x$ and $v=2y$. Then $u^2+v^2=4$ and $f=uv/2$. Since $$|uv|\le\frac{u^2+v^2}{2}=2,$$ we have $|f|\le 1$, with equality precisely when $|u|=|v|$. This reproduces all four points.