Lagrange Multipliers with One Constraint Examples 5

# Lagrange Multipliers with One Constraint Examples 5

Recall from The Method of Lagrange Multipliers page that with the Method of Lagrange Multipliers that if we have a function $z = f(x, y)$ subject to the constraint $g(x, y) = C$, then if extreme values exist and $\nabla g (x, y) \neq (0, 0)$ on the level curve $g(x, y) = C$ then we can obtain the critical points subject to the constraint by solving the system of equations below and then comparing the values of $f$ at these critical points.

(1)
\begin{align} \quad \nabla f(x, y) = \lambda \nabla g(x, y) \\ \quad g(x, y) = C \end{align}

Furthermore, if we have a function $w = f(x, y, z)$ subject to the constraint $g(x, y, z) = C$, then if extreme values exist and $\nabla g(x, y, z) \neq (0,0,0)$ on the level surface $g(x, y, z) = C$, then we can find these extreme values by solving the system of equations below and then comparing the values of $f$ at these critical points.

(2)
\begin{align} \quad \nabla f(x, y, z) = \lambda g(x, y, z) \\ \quad g(x, y, z) = C \end{align}

We will now look at some more examples of solving problems regarding Lagrange multipliers.

## Example 1

Find the extreme values of $f(x, y) = y$ subject to the constraint $y^3 - x^2 = 0$.

Let $g(x, y) = y^3 - x^2 = 0$. In computing the necessary partial derivatives we get that:

(3)
\begin{align} \quad 0 = \lambda ( -2x) \\ \quad 1 = \lambda (3y^2) \\ \quad y^3 - x^2 = 0 \end{align}

We first note that equation 1 implies that either $\lambda = 0$ or $x = 0$. We note that $\lambda \neq 0$ since this would contradict equation 2. Furthermore, $x \neq 0$ as well since if it did, equation 3 would imply that $y = 0$ which contradicts equation 2.

Thus, we note that there is no solution to the system of equations above. Recall that for the method of Lagrange multipliers to work, we needed that $\nabla g \neq 0$ on the level curve $g(x, y) = 0$. In particular, the gradient of $g$ is $\nabla g(x, y) = (-2x, 3y^2)$ which equals zero at the origin, however, the level curve $g(x, y) = y^3 - x^2 = 0$ passes through the origin.

Fortunately we can solve this problem with some substitution. Note that $y^3 - x^2 = 0$ implies that $y = x^{2/3}$. And thus, finding the extrema of $f$ constrained to $g$ is equivalent to finding the extrema of the single variable function $F$:

(4)
\begin{align} \quad f \left ( x, x^{2/3}\right ) = x^{2/3} = F(x) \end{align}

The graph of $F(x) = x^{2/3}$ is given below. Note that a cusp occurs at the origin where the gradient of $g$ is equal to zero. ## Example 2

Find the extreme values of $f(x_1, x_2, ..., x_n) = x_1 + x_2 + ... + x_n$ subject to the constraint equation $x_1^2 + x_2^2 + ... + x_n^2 = 1$.

We haven't yet discussed how to deal with Lagrange multiplier problems concerning functions of more than 3 variable, however, the intuition is exactly the same. We will want to solve the following system of equations

(5)
\begin{align} \quad 1 = 2\lambda x_1 \\ \quad 1 = 2\lambda x_2 \\ \quad \vdots \quad \\ \quad 1 = 2\lambda x_n \\ x_1^2 + x_2^2 + ... + x_n^2 = 1 \end{align}

We note that $x_1, x_2, ..., x_n \neq 0$ since this would lead to a contradict in the first $n$ equations above, and so we can freely divide by these variables to get that:

(6)
\begin{align} \lambda = \frac{1}{2x_1} = \frac{1}{2x_2} = ... = \frac{1}{2x_n} \end{align}

This implies that $2x_1 = 2x_2 = ... = 2x_n$ so $x_1 = x_2 = ... = x_n$, or equivalently, $x_j = x_1$ for $j = 2, 3, ..., n$, and so plugging this into the last equation yields:

(7)
\begin{align} \quad x_1^2 + x_1^2 + ... + x_1^2 = 1 \\ \quad nx_1^2 = 1 \\ \quad x_1^2 = \frac{1}{n} \\ \quad x_1 = \pm \frac{1}{\sqrt{n}} \end{align}

Therefore the points $\left ( \frac{1}{\sqrt{n}} , \frac{1}{\sqrt{n}} , ..., \frac{1}{\sqrt{n}} \right )$ and $\left ( -\frac{1}{\sqrt{n}} , -\frac{1}{\sqrt{n}} , ..., - \frac{1}{\sqrt{n}} \right )$ are of interest. The first point yields an extreme value of $\frac{n}{\sqrt{n}} = \sqrt{n}$ which is our maximum, while the second point yields an extreme value of $-\frac{n}{\sqrt{n}} = -\sqrt{n}$ which is our minimum.