Absolute Minimums and Maximums — Question 10

Question 10 – Solution

Strategy. A complete candidate search must include nondifferentiable points. On the circle, replace $y^2$ by $1-x^2$.

Step 1: Interior and crease Where $x\ne 0$, $$f_x=\operatorname{sgn}(x),\qquad f_y=2y,$$ so there are no differentiable interior critical points. On the crease $x=0$, $$f(0,y)=y^2.$$ Its smallest value is $0$ at the origin. Since $|x|\ge 0$ and $y^2\ge 0$ everywhere, $$\boxed{f_{\min}=0\text{ at }(0,0)}.$$

Step 2: Circular boundary Let $u=|x|\in[0,1]$. On $x^2+y^2=1$, $$f=u+1-u^2 =\frac 54-\left(u-\frac 12\right)^2.$$ Thus the largest boundary value is $5/4$, attained when $|x|=1/2$. Then $$y^2=1-\frac 14=\frac 34,$$ so $y=\pm\sqrt 3/2$.

Step 3: Final comparison For a fixed $x$, increasing $y^2$ increases $f$, so an absolute maximum must lie on the circle; the boundary calculation is therefore decisive. Hence $$\boxed{f_{\max}=\frac 54 \text{ at }\left(\pm\frac 12,\pm\frac{\sqrt 3}{2}\right)},$$ where the two signs are independent, giving four points. Direct substitution returns $1/2+3/4=5/4$ at each.