Lagrange Multipliers

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Objectives:

1.: Understand how the method of Lagrange multipliers works both algebraically and geometrically.
2.: Be comfortable finding critical points using the method of Lagrange multipliers.

Recap Video

The following video recaps the ideas of the section.

To recap:

Suppose $f(x,y)$ and $g(x,y)$ are differentiable functions. If $f(x,y)$ has a maximum or minimum on the constraint curve $g(x,y)= c$ (where $c$ is any number) at the point $P(a,b)$ , and if $\nabla g(a,b) \neq \vec {0}$ , then $\nabla f(a,b) = \lambda \nabla g(a,b),$ for some constant $\lambda$ (called the Lagrange multiplier).

Example Video

Below is a video working through an example of Lagrange multipliers.

Find the global maximum and global minimum of the function $f(x,y) = 2x^2+(y+1)^2$ on the disc $x^2+y^2 \leq 25$ .

Problems

Find the maximum and minimum values of the function $f(x,y) = x+y$ on the ellipse $x^2+2y^2 \leq 6$ .

First notice that this is a filled in ellipse (because it is $\leq 6$ , not $=6$ ). The region isis not closed and bounded, which means the global maximum and minimum values areare not guaranteed to exist. However, Lagrange multipliers only helps on the boundary, so we first need to deal with the inside of the ellipse.

Find critical points inside the ellipse: Do this by setting $\nabla f = \vec {0}$ . Here, $\nabla f = \left \langle \answer {1},\answer {1} \right \rangle .$ Can this ever equal $\vec {0}$ ?
Yes No
Therefore, there areare not critical points inside the region.
Deal with the boundary: The boundary is , which now looks like , where and . We now set . This means This gives us three equations: The goal is to accumulate all simultaneous solutions to these three equations. The multiplier is just there to help you in case you need it.
- Approach 1: One way to solve these equations is to solve for $\lambda$ in the first two equations. Looking at the first equation, can $x=0$ ? . Therefore, we can divide by $2x$ in the first equation, which gives $\lambda = \frac {1}{2x}$ . Looking at the first equation, can $y=0$ ? . Therefore, we can divide by $2y$ in the second equation to get $\lambda = \frac {1}{4y}$ . We can then get $x$ in terms of $y$ , and so we have $x = \answer {2y}$ . Plugging this into the constraint $x^2+2y^2=6$ gives $\answer {6y^2} = 6$ , so $y = \pm \answer {1}$ . Since $x = 2y$ , we get two points: $(\answer {2},1), \quad (\answer {-2},-1).$ These are all the critical points. Now plugging in these points into $f(x,y)$ shows that we have a maximum value of: $\answer {3}$ and a minimum value of: $\answer {-3}$ .
- Approach 2: We could also solve for $x$ and $y$ in terms of $\lambda$ if we wanted to. Looking at the first equation, can $\lambda = 0$ ? . Therefore, we can divide by $2\lambda$ in the first equation and $4\lambda$ in the second equation to get $x = \answer {\frac {1}{2\lambda }}, \quad y = \answer {\frac {1}{4\lambda }}.$ Plugging these into the constraint equation gives $6 = x^2+2y^2 = \answer {\frac {3}{8\lambda ^2}}.$ Solving for $\lambda$ gives $\lambda = \pm \answer {1/4}$ . Using our equations for $x$ and $y$ in terms of $\lambda$ give the points $x = \pm \frac {1}{2\cdot 1/4} = \pm 2$ and $y = \pm \frac {1}{4\cdot 1/4} = \pm 1$ . Therefore, our two critical points are $(2,1)$ and $(-2,-1)$ , as before. We can now plug these into $f(x,y)$ to find the same maximum and minimum values as in the other approach.

Find the maximum value of the function $f(x,y) = xy$ on the ellipse $3x^2+y^2=6$ .

Now, the region is just the ellipse, not filled in. The region is still closed and bounded, but now there is no interior to check. Therefore, Lagrange is enough to figure this out. The constraint looks like $g(x,y) = c$ , where $g(x,y) = \answer {3x^2+y^2}$ and $c=6$ . We set $\nabla f = \lambda \nabla g$ , or $\left \langle \answer {y},\answer {x} \right \rangle = \lambda \left \langle \answer {6x},\answer {2y} \right \rangle .$ This gives us three equations $\begin{equation} \label{eq:1} y= \lambda \answer{6x} \end{equation}$ $\begin{equation} \label{eq:2} x = \lambda \answer{2y}\end{equation}$ $\begin{equation} \label{eq:3} 3x^2+y^2=6.\end{equation}$ You may already have your own way for solving this, but we will show you two approaches.

Approach 1: We can solve for $\lambda$ in each of the first two equations. As in the previous problem, we should check whether the variables can be $0$ before dividing. For the first equation, if $x=0$ , then we see $y = \answer {0}$ , which work in the last equation. This tells us that $x$ equal $0$ . Following the same logic for the second equation tells us $y$ cannot be $0$ . Therefore, we can divide the first equation by $6x$ and second by $2y$ to get $\lambda = \frac {y}{6x} = \frac {x}{2y}.$ Cross multiplying and simplifying gives us $y^2 = \answer {3x^2}.$ We can now plug this into our constraint to get $3x^2+\answer {3x^2} = 6.$ Solving for $x$ gives $x = \pm \answer {1}$ . Since $y^2 = 3x^2$ , we can solve for $y$ with each of these $x$ values, giving us a total of four critical points. The critical points are $(1,\pm \answer {\sqrt {3}}),\quad (-1,\pm \answer {\sqrt {3}}).$
Approach 2: We could also just multiply the first equation by $x$ to get $xy = 6x^2\lambda$ and second by $y$ to get $xy =\answer {2y^2}\lambda .$ Therefore $6x^2\lambda = 2y^2\lambda$ . If $\lambda = 0$ , then equation eq:1 tells us $y = \answer {0}$ and equation eq:2 tells us $x=\answer {0}$ , which again does not work in in equation eq:3. Therefore, we can divide by $\lambda$ to get $6x^2 = 2y^2$ , or $y^2 = 3x^2$ . Now we repeat the steps from Approach 1 above to get the four critical points.

Plugging in all four points into $f(x,y)$ gives us a maximum value of $\answer {\sqrt {3}}$ .

Find the maximum and minimum values of the function $f(x,y,z) = x+y+2z$ on the sphere $x^2+y^2+z^2=6$ .
The maximum value is $\answer {6}$ and the minimum value is: $\answer {-6}$ .

Find the maximum and minimum value of the function $f(x,y) = x^2y$ on the circle $x^2+y^2=1$ using Lagrange multipliers.
The maximum is: $\answer {\frac {2}{3\sqrt {3}}}$ and the minimum is: $\answer {-\frac {2}{3\sqrt {3}}}$ .

There is a method which is slightly shorter than Lagrange multipliers to solve the above problem. Can you find it?

Find the maximum and minimum value of $f(x,y,z) = x^2-y^2$ on the ellipsoid $x^2+2y^2+3z^2=1$ .
The maximum is $\answer {1}$ and the minimum is $\answer {-1/2}$