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## Multivariable calculus

### Course: Multivariable calculus>Unit 3

Lesson 6: Constrained optimization (articles)

# Lagrange multipliers, examples

Examples of the Lagrangian and Lagrange multiplier technique in action.

## Lagrange multiplier technique, quick recap

Constrained optimization
Image credit: By Nexcis (Own work) [Public domain], via Wikimedia Commons
When you want to maximize (or minimize) a multivariable function start color #0c7f99, f, left parenthesis, x, comma, y, comma, dots, right parenthesis, end color #0c7f99 subject to the constraint that another multivariable function equals a constant, start color #bc2612, g, left parenthesis, x, comma, y, comma, dots, right parenthesis, equals, c, end color #bc2612, follow these steps:
• Step 1: Introduce a new variable start color #0d923f, lambda, end color #0d923f, and define a new function L as follows:
L, left parenthesis, x, comma, y, comma, dots, comma, start color #0d923f, lambda, end color #0d923f, right parenthesis, equals, start color #0c7f99, f, left parenthesis, x, comma, y, comma, dots, right parenthesis, end color #0c7f99, minus, start color #0d923f, lambda, end color #0d923f, left parenthesis, start color #bc2612, g, left parenthesis, x, comma, y, comma, dots, right parenthesis, minus, c, end color #bc2612, right parenthesis
This function L is called the "Lagrangian", and the new variable start color #0d923f, lambda, end color #0d923f is referred to as a "Lagrange multiplier"
• Step 2: Set the gradient of L equal to the zero vector.
del, L, left parenthesis, x, comma, y, comma, dots, comma, start color #0d923f, lambda, end color #0d923f, right parenthesis, equals, start bold text, 0, end bold text, left arrow, start color gray, start text, Z, e, r, o, space, v, e, c, t, o, r, end text, end color gray
In other words, find the critical points of L.
• Step 3: Consider each solution, which will look something like left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, comma, dots, comma, start color #0d923f, lambda, end color #0d923f, start subscript, 0, end subscript, right parenthesis. Plug each one into f. Or rather, first remove the start color #0d923f, lambda, end color #0d923f, start subscript, 0, end subscript component, then plug it into f, since f does not have start color #0d923f, lambda, end color #0d923f as an input. Whichever one gives the greatest (or smallest) value is the maximum (or minimum) point your are seeking.

## Example 1: Budgetary constraints

#### Problem

Suppose you are running a factory, producing some sort of widget that requires steel as a raw material. Your costs are predominantly human labor, which is dollar sign, 20 per hour for your workers, and the steel itself, which runs for dollar sign, 170 per ton. Suppose your revenue R is loosely modeled by the following equation:
R, left parenthesis, h, comma, s, right parenthesis, equals, 200, h, start superscript, 2, slash, 3, end superscript, s, start superscript, 1, slash, 3, end superscript
• h represents hours of labor
• s represents tons of steel
If your budget is dollar sign, 20, comma, 000, what is the maximum possible revenue?

#### Solution

The dollar sign, 20 per hour labor costs and dollar sign, 170 per ton steel costs tell us that the total cost of production, in terms of h and s, is
\begin{aligned} \quad 20h + 170s \end{aligned}
Therefore the budget of dollar sign, 20, comma, 000 can be translated to the constraint
\begin{aligned} \quad \redE{20h + 170s = 20{,}000} \end{aligned}
Before we dive into the computation, you can get a feel for this problem using the following interactive diagram. You can see which values of left parenthesis, h, comma, s, right parenthesis yield a given revenue (blue curve) and which values satisfy the constraint (red line).
Since we need to maximize a function start color #0c7f99, R, left parenthesis, h, comma, s, right parenthesis, end color #0c7f99, subject to a constraint, start color #bc2612, 20, h, plus, 170, s, equals, 20, comma, 000, end color #bc2612, we begin by writing the Lagrangian function for this setup:
L, left parenthesis, h, comma, s, comma, lambda, right parenthesis, equals, start color #0c7f99, 200, h, start superscript, 2, slash, 3, end superscript, s, start superscript, 1, slash, 3, end superscript, end color #0c7f99, minus, lambda, left parenthesis, start color #bc2612, 20, h, plus, 170, s, minus, 20, comma, 000, end color #bc2612, right parenthesis
Next, set the gradient del, L equal to the start bold text, 0, end bold text vector. This is the same as setting each partial derivative equal to 0. First, we handle the partial derivative with respect to start color #0c7f99, h, end color #0c7f99.
\begin{aligned} \quad 0 &= \dfrac{\partial \mathcal{L}}{\partial \blueE{h}} \\ \\ 0 &= \dfrac{\partial}{\partial \blueE{h}}(200\blueE{h}^{2/3}s^{1/3}-\lambda(20\blueE{h}+170s-20{,}000)) \\ \\ 0 &= 200 \cdot \dfrac{2}{3}\blueE{h}^{-1/3}s^{1/3} - 20\lambda \\ \end{aligned}
Next, we handle the partial derivative with respect to start color #0d923f, s, end color #0d923f.
\begin{aligned} \quad 0 &= \dfrac{\partial \mathcal{L}}{\partial \greenE{s}} \\ \\ 0 &= \dfrac{\partial}{\partial \greenE{s}}(200h^{2/3}\greenE{s}^{1/3}-\lambda(20h+170\greenE{s}-20{,}000)) \\ \\ 0 &= 200 \cdot \dfrac{1}{3}h^{2/3}\greenE{s}^{-2/3} - 170\lambda \end{aligned}
Finally we set the partial derivative with respect to start color #a75a05, lambda, end color #a75a05 equal to 0, which as always is just the same thing as the constraint. In practice, you can of course just write the constraint itself, but I'll write out the partial derivative here just to make things clear.
\begin{aligned} \quad 0 &= \dfrac{\partial \mathcal{L}}{\partial \goldE{\lambda}} \\ \\ 0 &= \dfrac{\partial}{\partial \goldE{\lambda}}(200h^{2/3}s^{1/3}-\goldE{\lambda}(20h+170s-20{,}000)) \\ \\ 0 &= -20h-170s+20{,}000 \\ \\ 20h &+ 170s = 20{,}000 \end{aligned}
Putting it together, the system of equations we need to solve is
\begin{aligned} \quad 0 &= 200 \cdot \dfrac{2}{3}{h}^{-1/3}s^{1/3} - 20\lambda \\ \\ 0 &= 200 \cdot \dfrac{1}{3}h^{2/3}{s}^{-2/3} - 170\lambda \\ \\ 20h &+ 170s = 20{,}000 \end{aligned}
In practice, you should almost always use a computer once you get to a system of equations like this. Especially because the equation will likely be more complicated than these in real applications. Once you do, you'll find that the answer is
\begin{aligned} \quad h &= \dfrac{2{,}000}{3} \approx 666.667 \\ \\ s &= \dfrac{2{,}000}{51} \approx 39.2157 \\ \\ \lambda &= \sqrt{\dfrac{8{,}000}{459}} \approx 2.593 \\ \\ \end{aligned}
This means you should employ about 667 hours of labor, and purchase 39 tons of steel, which will give a maximum revenue of
\begin{aligned} \quad R(667, 39) = 200(667)^{2/3}(39)^{1/3} \approx \boxed{\51{,}777} \end{aligned}
The interpretation of this constant lambda, equals, 2, point, 593 is left to the next article

## Example 2: Maximizing dot product

Problem: Let the three-dimensional vector start bold text, v, end bold text, with, vector, on top be defined as follows.
\begin{aligned} \quad \vec{\textbf{v}} = \left[ \begin{array}{c} 2 \\ 3 \\ 1 \end{array} \right] \end{aligned}
Consider every possible unit vector start bold text, u, end bold text, with, hat, on top in three-dimensional space. For which one is the dot product start bold text, u, end bold text, with, hat, on top, dot, start bold text, v, end bold text, with, vector, on top the greatest?
The diagram below is two-dimensional, but not much changes in the intuition as we move to three dimensions.
Unit vector dot product
Two-dimensional analogy to the three-dimensional problem we have. Which unit vector start bold text, u, end bold text, with, hat, on top maximizes the dot product start bold text, u, end bold text, with, hat, on top, dot, start bold text, v, end bold text, with, vector, on top?
If you are fluent with dot products, you may already know the answer. It's one of those mathematical facts worth remembering. If you don't know the answer, all the better! Because we will now find and prove the result using the Lagrange multiplier method.
Solution:
First, we need to spell out how exactly this is a constrained optimization problem. Write the coordinates of our unit vectors as x, y and z:
\begin{aligned} \quad \hat{\textbf{u}} = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] \end{aligned}
The fact that start bold text, u, end bold text, with, hat, on top is a unit vector means its magnitude is 1:
\begin{aligned} \quad ||\hat{\textbf{u}}|| = \sqrt{x^2 + y^2 + z^2} &= 1 \\ &\Downarrow \\ \redE{x^2 + y^2 + z^2} &\redE{= 1} \end{aligned}
This is our constraint.
Maximizing start bold text, u, end bold text, with, hat, on top, dot, start bold text, v, end bold text, with, vector, on top means maximizing the following quantity:
$\left[ \begin{array}{c} x \\ y \\ z \end{array} \right] \cdot \left[ \begin{array}{c} 2 \\ 3 \\ 1 \end{array} \right] = \blueE{2x + 3y + z}$
The Lagrangian, with respect to this function and the constraint above, is
L, left parenthesis, x, comma, y, comma, z, comma, lambda, right parenthesis, equals, 2, x, plus, 3, y, plus, z, minus, lambda, left parenthesis, x, squared, plus, y, squared, plus, z, squared, minus, 1, right parenthesis, point
We now solve for del, L, equals, start bold text, 0, end bold text by setting each partial derivative of this expression equal to 0.
\begin{aligned} \dfrac{\partial}{\partial \greenE{x}}(2\greenE{x} + 3y + z - \lambda(\greenE{x}^2 + y^2 + z^2 - 1)) &= 2 - \lambda 2\greenE{x} = 0 \\ \dfrac{\partial}{\partial \goldE{y}}(2x + 3\goldE{y} + z - \lambda(x^2 + \goldE{y}^2 + z^2 - 1)) &= 3 - \lambda 2\goldE{y} = 0 \\ \dfrac{\partial}{\partial \maroonE{z}}(2x + 3y + \maroonE{z} - \lambda(x^2 + y^2 + \maroonE{z}^2 - 1)) &= 1 - \lambda 2\maroonE{z} = 0 \\ \end{aligned}
Remember, setting the partial derivative with respect to lambda equal to 0 just restates the constraint.
\begin{aligned} \dfrac{\partial}{\partial \redE{\lambda}}(2x + 3y + z - \redE{\lambda}(x^2 + y^2 + z^2 - 1)) &= -x^2-y^2-z^2 + 1 = 0 \\ \end{aligned}
Solving for start color #0d923f, x, end color #0d923f, start color #a75a05, y, end color #a75a05 and start color #9e034e, z, end color #9e034e in the first three equations above, we get
\begin{aligned} \quad \greenE{x} &= 2\cdot \dfrac{1}{2\lambda} \\ \goldE{y} &= 3\cdot \dfrac{1}{2\lambda} \\ \maroonE{z} &= 1\cdot \dfrac{1}{2\lambda} \end{aligned}
Ah, what beautiful symmetry. Each of these expressions has the same start fraction, 1, divided by, 2, lambda, end fraction factor, and the coefficients 2, 3 and 1 match up with the coordinates of start bold text, v, end bold text, with, vector, on top. Being good math students as we are, we won't let good symmetry go to waste. In this case, combining the three equations above into a single vector equation, we can relate start bold text, u, end bold text, with, hat, on top and start bold text, v, end bold text, with, vector, on top as follows:
\begin{aligned} \quad \hat{\textbf{u}} = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] = \dfrac{1}{2\lambda} \left[ \begin{array}{c} 2 \\ 3 \\ 1 \end{array} \right] = \dfrac{1}{2\lambda}\vec{\textbf{v}} \end{aligned}
Two-dimensional analogy showing the two unit vectors which maximize and minimize the quantity start bold text, u, end bold text, with, hat, on top, dot, start bold text, v, end bold text, with, vector, on top.
Therefore start bold text, u, end bold text, with, hat, on top is proportional to start bold text, v, end bold text, with, vector, on top! Geometrically, this means start bold text, u, end bold text, with, hat, on top points in the same direction as start bold text, v, end bold text, with, vector, on top. There are two unit vectors proportional start bold text, v, end bold text, with, vector, on top,
• One which points in the same direction, this is the vector that start color #0d923f, start text, m, a, x, i, m, i, z, e, s, end text, end color #0d923f start bold text, u, end bold text, with, hat, on top, dot, start bold text, v, end bold text, with, vector, on top.
• One which points in the opposite direction. This one start color #bc2612, start text, m, i, n, i, m, i, z, e, s, end text, end color #bc2612 start bold text, u, end bold text, with, hat, on top, dot, start bold text, v, end bold text, with, vector, on top.
We can write these two unit vectors by normalizing start bold text, v, end bold text, with, vector, on top, which just means dividing start bold text, v, end bold text, with, vector, on top by its magnitude:
\begin{aligned} \quad \greenE{\hat{\textbf{u}}_{\text{max}}} &= \dfrac{\vec{\textbf{v}}}{||\vec{\textbf{v}}||} \\ \\ \redE{\hat{\textbf{u}}_{\text{min}}} &= -\dfrac{\vec{\textbf{v}}}{||\vec{\textbf{v}}||} \end{aligned}
The magnitude vertical bar, vertical bar, start bold text, v, end bold text, with, vector, on top, vertical bar, vertical bar is square root of, 2, squared, plus, 3, squared, plus, 1, squared, end square root, equals, square root of, 14, end square root, so we can write the maximizing unit vector start color #0d923f, start bold text, u, end bold text, with, hat, on top, start subscript, start text, m, a, x, end text, end subscript, end color #0d923f explicitly as like this:
$\greenE{\hat{\textbf{u}}_{\text{max}}} = \left[ \begin{array}{c} 2 / \sqrt{14} \\ 3 / \sqrt{14} \\ 1 / \sqrt{14} \end{array} \right]$

## Just skip the Lagrangian

If you read the last article, you'll recall that the whole point of the Lagrangian L is that setting del, L, equals, 0 encodes the two properties a constrained maximum must satisfy:
• Gradient alignment between the target function and the constraint function,
\begin{aligned} \quad \nabla \blueE{f(x, y)} &= \lambda \nabla \redE{g(x, y)} \\ \end{aligned}
• The constraint itself,
\begin{aligned} \quad \redE{g(x, y) = c} \end{aligned}
When working through examples, you might wonder why we bother writing out the Lagrangian at all. Wouldn't it be easier to just start with these two equations rather than re-establishing them from del, L, equals, 0 every time? The short answer is yes, it would be easier. If you find yourself solving a constrained optimization problem by hand, and you remember the idea of gradient alignment, feel free to go for it without worrying about the Lagrangian.
In practice, it's often a computer solving these problems, not a human. Given that there are many highly optimized programs for finding when the gradient of a given function is 0, it's both clean and useful to encapsulate our problem into the equation del, L, equals, 0.
Furthermore, the Lagrangian itself, as well as several functions deriving from it, arise frequently in the theoretical study of optimization. In this light, reasoning about the single object L rather than multiple conditions makes it easier to see the connection between high-level​ ideas. Not to mention, it's quicker to write down on a blackboard.
In either case, whatever your future relationship with constrained optimization might be, it is good to be able to think about the Lagrangian itself and what it does. The examples above illustrate how it works, and hopefully help to drive home the point that del, L, equals, 0 encapsulates both del, f, equals, lambda, del, g and g, left parenthesis, x, comma, y, right parenthesis, equals, c in a single equation.