How to Use Lagrange Multipliers

Last Updated: May 3, 2022

In calculus, Lagrange multipliers are commonly used for constrained optimization problems. These types of problems have wide applicability in other fields, such as economics and physics. The basic structure of a Lagrange multiplier problem is of the relation below:

${\mathcal {L}}(x,y;\lambda )=f(x,y)+\lambda g(x,y)$

where $f(x,y)$ is the function to be optimized, $g(x,y)$ is the constraint, and $\lambda$ is the Lagrange multiplier. Then, we set $\nabla {\mathcal {L}}=\nabla f+\lambda \nabla g=0$ to solve the resulting system of equations; oftentimes, we wish to cancel out $\lambda$ in the process. These problems can easily be generalized to higher dimensions and more constraints.

Part 1

Part 1 of 2:

Example 1

1
Find the maximum value of $x^{3}y$ on the ellipse $3x^{2}+y^{2}=6$ . This is a Lagrange multiplier problem, because we wish to optimize a function subject to a constraint. In optimization problems, we typically set the derivatives to 0 and go from there. But in this case, we cannot do that, since the max value of $x^{3}y$ $x^{{3}}y$ may not lie on the ellipse.
- Clearly, $f(x,y)=x^{3}y$ and $g(x,y)=3x^{2}+y^{2}=6.$
2
Take the gradient of the Lagrangian ${\mathcal {L}}$ . Setting it to 0 gets us a system of two equations with three variables.
- $\nabla f+\lambda \nabla g=0$
- ${\begin{cases}3x^{2}y+\lambda 6x&=0\\x^{3}+\lambda 2y&=0\end{cases}}$
Advertisement
3
Cancel $\lambda$ and set the equations equal to each other. Since we are not concerned with it, we need to cancel it out. Here, we multiply the first equation by $y$ $y$ and the second equation by $3x.$ $3x.$
- ${\begin{cases}3x^{2}y^{2}+\lambda 6xy&=0\\3x^{4}+\lambda 6xy&=0\end{cases}}$
- ${\begin{aligned}3x^{2}y^{2}-3x^{4}&=0\\x^{2}(y^{2}-x^{2})&=0\end{aligned}}$
4
Relate $x$ with $y$ . In the above equation, we see that when $x\neq 0,\ y^{2}-x^{2}=0.$ $x\neq 0,\ y^{{2}}-x^{{2}}=0.$ This gets us the relation below.
- $y=x$
5
Substitute the expression for $y$ in terms of $x$ into the constraint equation. Now that we have derived this useful relation, we can finally find values for $x$ $x$ and $y.$ $y.$
- ${\begin{aligned}g&=3x^{2}+x^{2}=6\\x&=y=\pm {\sqrt {\frac {3}{2}}}\end{aligned}}$
6
Substitute the values of $x$ and $y$ into the optimization equation. We have found the maximum value of the function $x^{3}y$ $x^{{3}}y$ on the ellipse $3x^{2}+y^{2}=6.$ $3x^{{2}}+y^{{2}}=6.$
- $x^{3}y={\frac {9}{4}}$

Part 2

Part 2 of 2:

Example 2

1
Find the minimum distance from $x^{2}yz=1$ to the origin. Recall the distance as ${\sqrt {x^{2}+y^{2}+z^{2}}}.$ ${\sqrt {x^{{2}}+y^{{2}}+z^{{2}}}}.$ This is the function that we are trying to optimize, with the constraint function as $x^{2}yz-1=0.$ $x^{{2}}yz-1=0.$ This is a somewhat difficult expression to work with, however. In this case, we can remove the square root and optimize $x^{2}+y^{2}+z^{2}$ $x^{{2}}+y^{{2}}+z^{{2}}$ instead, since we are working in the same domain (only positive numbers), so the numbers will turn out to be the same. We just have to remember that the function to be optimized is the expression with the square root.
- ${\mathcal {L}}(x,y,z;\lambda )=(x^{2}+y^{2}+z^{2})+\lambda (x^{2}yz-1)$
2
Take the gradient of the Lagrangian and set each component to 0.
- ${\begin{cases}{\frac {\partial {\mathcal {L}}}{\partial x}}=2x+\lambda 2xyz&=0\\{\frac {\partial {\mathcal {L}}}{\partial y}}=2y+\lambda x^{2}z&=0\\{\frac {\partial {\mathcal {L}}}{\partial z}}=2z+\lambda x^{2}y&=0\end{cases}}$
3
Cancel out $\lambda$ . Here, multiply the first equation by $x,$ $x,$ the second equation by $2y,$ $2y,$ and the third equation by $2z.$ $2z.$
- ${\begin{cases}2x^{2}+\lambda 2x^{2}yz&=0\\4y^{2}+\lambda 2x^{2}yz&=0\\4z^{2}+\lambda 2x^{2}yz&=0\end{cases}}$
4
Relate the variables to each other by solving for one of them. Let's use $y,$ $y,$ though $x$ $x$ and $z$ $z$ are fine too.
- $x^{2}=2y^{2}=2z^{2}$
- The equation above gives us all the information we need to optimize the distance now.
5
Obtain the value for $y$ by substituting into the constraint function. Since we know $y=z,$ $y=z,$ we can write the constraint function in terms of just $y$ $y$ and solve for it.
- ${\begin{aligned}(2y^{2})(y)(y)&=1\\y&=2^{-1/4}\end{aligned}}$
6
Substitute the value for $y$ into the distance. Remember, even though we were optimizing the square of the distance, we are still looking for the actual distance.
- ${\begin{aligned}{\sqrt {x^{2}+y^{2}+z^{2}}}&={\sqrt {2y^{2}+y^{2}+y^{2}}}\\&={\sqrt {4y^{2}}}\\&=2\cdot 2^{-1/4}\\&=2^{3/4}\end{aligned}}$