Exercise 734Calculus › Multivariable calculus › Optimisation in several variablesL2Added 2 Nov 2026

Optimisation under a linear constraint

#Lagrange#linear constraint#minimum

Statement

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

Hint

Lagrange with $g(x,y) = x+2y-5$ . The Lagrangian $L = x^2+y^2 - \lambda(x+2y-5)$ is convex, so the critical point is a global minimum.

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Solution

Lagrange system.
$\begin{cases} 2x = \lambda \\ 2y = 2\lambda \\ x + 2y = 5 \end{cases}$

Resolution. From the first equation, $\lambda = 2x$ . From the second, $2y = 4x$ , so $y = 2x$ . Substituting into the constraint:
$x + 4x = 5 \implies 5x = 5 \implies x = 1, \quad y = 2.$

Conclusion. The minimum is at $(1,2)$ , with $f(1,2) = 1 + 4 = 5$ . It is indeed a global minimum because the Lagrangian is convex (sum of squares and a linear form).

Optimisation under a linear constraint

Statement

Solution

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