Exercise 1592Calculus › Multivariable calculus › Optimisation in several variablesL1Added 3 Nov 2026

Optimization under inequality constraints

#inequality constraint#KKT#boundary

Statement

Maximize $f(x,y)=x+y$ on the disk $x^2+y^2\le1$ .

Hint

Interior critical points or boundary via Lagrange.

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Interior. $\nabla f=(1,1)\neq(0,0)$ : no interior critical point.

Boundary. Lagrange on $x^2+y^2=1$ : $1=2\lambda x$ , $1=2\lambda y\Rightarrow x=y$ , so $x=y=\pm1/\sqrt2$ .

Values. $f(1/\sqrt2,1/\sqrt2)=\sqrt2$ (max); $f(-1/\sqrt2,-1/\sqrt2)=-\sqrt2$ (min).

Conclusion. Maximum $\sqrt2$ at $(1/\sqrt2,1/\sqrt2)$ .