Exercise 860Calculus › Multivariable calculus › Optimisation in several variablesL1Added 3 Nov 2026

Lagrange multipliers on a circle

#Lagrange#circle#extrema

Statement

Find the extrema of $f(x,y)=3x-y$ under the constraint $x^2+y^2=5$ .

Hint

$\nabla f=\lambda\nabla g$ with $g=x^2+y^2-5$ .

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Lagrange system.
$3=2\lambda x,\quad -1=2\lambda y,\quad x^2+y^2=5.$

Solve. $x=3/(2\lambda)$ , $y=-1/(2\lambda)$ , so $y=-x/3$ . Then $x^2+x^2/9=5\Rightarrow x^2=9/2\Rightarrow x=\pm3/\sqrt2$ .

Points. $A=(3/\sqrt2,-1/\sqrt2)$ , $B=(-3/\sqrt2,1/\sqrt2)$ .

Values. $f(A)=5\sqrt2$ (maximum), $f(B)=-5\sqrt2$ (minimum).

Conclusion. Maximum $5\sqrt2$ at $A$ , minimum $-5\sqrt2$ at $B$ .