# Extrema of multivalued function with constraint

This is not supposed to be a 'how-to-do' question, but rather a 'why' question. I came across the following problem:

$Let\ f: R^2 \rightarrow R;\ \ f(x,y)=x^3+y^3.\ Find\ the\ extrema\ points\ of\ f\ so\ that\ x^2+y^2=1$

The solution for the minimum works in the same way as that for finding the maximum, so I'll just consider the maximum when explaining my approach. I took the Lagrange function, made its partial derivatives equal to $0,\$ found $\ \lambda\$ so that $g(x,y)=x^2+y^2-1=0\$ which resulted in some possible solutions. What I find strange is when checking the solution of this problem $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\$ I found given as a maximum point, whereas $(1,0)$ (and, of course, $(0,1))\$ clearly yields a greater $\ f$ while also verifying $g(x,y)=0$ and all the partial differential tests for $\ \lambda=-\frac{3}{2}$. Still, $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\$ passes the Hessian test while $(1,0)$ doesn't, but this just leaves the Hessian inconclusive for $(1,0)$, which shouldn't be a problem.

So which is the maximum of $\ f$ on $\ x^2+y^2=1?$ $$(1,0)\ and\ (0,1)\ or\ (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})?$$

with $$y=\pm\sqrt{1-x^2}$$ you will get $$f(x,\pm\sqrt{1-x^2})=x^3+(\pm\sqrt{1-x^2})^3$$ and you Problem is reduced to a Problem in only one variable.

• I understand. So, how come there is no mention of (1,0) in my textbook? With your method, it clearly is a maximum point and since f(1,0) > f(sqrt(2)/2,sqrt(2)/2), (1,0) should be the actual global maximum. The second derivative test also indicates nothing in the case of (1,0), but why would this "disqualify" (1,0) as a maximum point? Or does it? – Escu Esculescu Jan 18 '18 at 19:07
• $1$ is clearly the Maximum for $x=0,y=1$ – Dr. Sonnhard Graubner Jan 18 '18 at 19:36