Lagrange Multipliers
Lagrange Multipliers
To optimize $f(x,y)$ subject to the constraint $g(x,y) = 0$, solve:
$$\nabla f = \lambda \nabla g \quad \text{and} \quad g(x,y) = 0$$
This gives the system: $f_x = \lambda g_x$, $f_y = \lambda g_y$, $g = 0$.
At a constrained extremum, $\nabla f$ is parallel to $\nabla g$ — the level curves of $f$ are tangent to the constraint curve.
Maximize $f(x,y) = xy$ subject to $x + y = 10$.
$g(x,y) = x + y - 10 = 0$. $\nabla f = \langle y, x \rangle$, $\nabla g = \langle 1, 1 \rangle$.
$y = \lambda$, $x = \lambda$ → $x = y$. From constraint: $2x = 10$, $x = y = 5$.
$f(5,5) = 25$.
Minimize $f(x,y) = x^2 + y^2$ subject to $x + 2y = 5$.
$2x = \lambda$, $2y = 2\lambda$ → $y = 2x$. $x + 4x = 5$, $x = 1$, $y = 2$.
$f(1,2) = 5$.
Maximize $f(x,y,z) = xyz$ subject to $x + y + z = 12$, $x,y,z > 0$.
$yz = \lambda$, $xz = \lambda$, $xy = \lambda$ → $x = y = z$. $3x = 12$, $x = y = z = 4$.
$f = 64$.
Practice Problems
Show Answer Key
1. $1 = 2\lambda x$, $1 = 2\lambda y$ → $x=y$. $2x^2=1$; $(1/\sqrt{2},1/\sqrt{2})$. $f=\sqrt{2}$.
2. $x=y=2$. $f=8$.
3. $y=2\lambda x$, $x=2\lambda y$ → $x^2=y^2$, $x=y=2$. $f=4$.
4. $2=2\lambda x$, $1=2\lambda y$ → $y=x/2$. $x^2+x^2/4=5$; $x=2,y=1$. $f=5$. Min when $x=-2,y=-1$: $f=-5$.
5. Marginal value — rate of change of optimal value per unit relaxation of constraint
6. $2xy=2\lambda$, $x^2=\lambda$. $2xy=2x^2$ → $y=x$. $3x=6$; $x=y=2$... actually $y=6-2x$: use $2xy=2\lambda$, $x^2=\lambda$ → $y = x$. $2x+x=6$; $x=2,y=2$. $f=8$.
7. Minimize $x^2+y^2$ with $3x+4y=25$. $(3,4)$. Distance $= 5$.
8. $x=y=z=1$. $f=3$.
9. $3$ ($f_x=\lambda g_x$, $f_y=\lambda g_y$, $g=0$)
10. $x=y=z=2$. $f=12$.
11. $2$ ($\lambda_1$ and $\lambda_2$)
12. $y=2\lambda(3)=6\lambda$, $x=3\lambda(2)=... $ Use $y=2\lambda$, $x=3\lambda$... actually $y=2\lambda$, $x=3\lambda$. $2(3\lambda)+3(2\lambda)=12$; $12\lambda=12$; $\lambda=1$. $x=3, y=2$. $f=6$.