Exercise 861Calculus › Multivariable calculus › Optimisation in several variablesL1Added 3 Nov 2026

Optimization under a linear constraint

#Lagrange#linear constraint#minimum

Statement

Minimize $f(x,y)=x^2+y^2$ under $x+2y=5$ .

Hint

Substitute $y=(5-x)/2$ or use Lagrange.

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Lagrange. $2x=\lambda$ , $2y=2\lambda\Rightarrow y=2x$ ; constraint gives $5x=5$ , $x=1$ , $y=2$ .

Value. $f(1,2)=5$ .

Conclusion. Minimum $5$ at $(1,2)$ . The Lagrangian is convex so the critical point is a global minimum.