Absolute Minimums and Maximums — Question 4

Question 4 – Solution

Strategy. Let $u=x$ and $v=2y$; then the region is a radius-$2$ disk and the objective is $u+v$.

Step 1: Transform Set $$u=x,\qquad v=2y.$$ The constraint becomes $u^2+v^2\le 4$, while $$f=u+v=\left\langle u,v\right\rangle\cdot\left\langle 1,1\right\rangle.$$

Step 2: Sharp bounds By Cauchy–Schwarz, $$|u+v|\le\sqrt{u^2+v^2}\sqrt 2\le 2\sqrt 2.$$ Thus $-2\sqrt 2\le f\le 2\sqrt 2$.

Equality on the upper side requires $\left\langle u,v\right\rangle$ to point with $\left\langle 1,1\right\rangle$ and have length $2$, giving $$(u,v)=(\sqrt 2,\sqrt 2).$$ Lower equality gives its negative.

Step 3: Return to $(x,y)$ Because $x=u$ and $y=v/2$, $$\boxed{f_{\max}=2\sqrt 2\text{ at }\left(\sqrt 2,\frac{\sqrt 2}{2}\right)},$$ $$\boxed{f_{\min}=-2\sqrt 2\text{ at }\left(-\sqrt 2,-\frac{\sqrt 2}{2}\right)}.$$ For either point, $x^2+4y^2=2+2=4$, verifying boundary membership and sharpness.