Lagrange Multipliers with One Constraint Examples 1

# Lagrange Multipliers with One Constraint Examples 1

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 maximum and minimum values of the function $f(x, y) = 2x^2 + 3y^2$ subject to the constraint $2x + 2y = 1$ using Lagrange multipliers and by direct substitution.

Let $g(x, y) = 2x + 2y = 1$. Computing the necessary partial derivatives and we have that:

(3)
\begin{align} \quad 4x = 2 \lambda \\ \quad 6y = 2 \lambda \\ 2x + 2y = 1 \end{align}

From the first equation, we immediately have that $4x = 2 \lambda$ implies that $\lambda = 2x$. Plugging this into the second equation will eliminate one of the equations and thus we have that $6y = 2 \lambda = 2(2x) = 4x$. Thus we have that:

(4)
\begin{align} \quad y = \frac{2}{3}x \\ \quad 2x + 2y = 1 \end{align}

Plugging the first equation into the second equation and we have that $2x + 2 \left (\frac{2}{3}x \right ) = 1$. Therefore $2x + \frac{4}{3}x = 1$, so $\frac{10}{3}x = 1$ and $x = \frac{3}{10}$. Plugging this back into the first equation and we get that $y = \frac{1}{5}$. Therefore $\left ( \frac{3}{10}, \frac{1}{5} \right )$ is a point of interest.

Now evaluating $f$ at this point of interest and we get that:

(5)
\begin{align} \quad f \left ( \frac{3}{10}, \frac{1}{5} \right ) = \frac{3}{10} \end{align}

Now note that if $x$ gets arbitrarily large positively, then $y$ can get arbitrarily large negatively, but their squares get arbitrarily large. Thus $\frac{3}{10}$ is a minimum of $f$ under the constraint imposed by $g$.

The following image is the graph of the ellipsoid $f(x, y) = 2x^2 + 3y^2$. The yellow curve represents the points for which our maximum/minimum problem is constrained. The blue dot represents the minimum of $f$ subject to the constraint $g$. Furthermore, the following image is a graph of the level curves of $f$ alongside the curve $g(x, y) = 1$. The blue point represents our point of interest, $\left ( \frac{3}{10}, \frac{1}{5} \right )$ in the domain of $f$. Now we will use direct substitution. Noting that the constraint equation can be rewritten as $y = \frac{1 - 2x}{2} = \frac{1}{2} - x$. Substituting this for $y$ in the function $f$ and we get a function of a single variable:

(6)
\begin{align} \quad 2x^2 + 3y^2 = 2x^2 + 3 \left ( \frac{1}{2} - x \right )^2 = 2x^2 + 3 \left ( \frac{1}{4} - x + x^2 \right ) = 5x^2 - 3x + \frac{3}{4} = f(x) \end{align}

This function is a positive parabola, and so it obtained a minimum value at its vertex. The vertex for this parabola is $\frac{-b}{2a}$ where $b = -3$ and $a = 5$, thus the vertex is located at $x = \frac{3}{10}$. Plugging this value of $x$ into our constraint equation and we get that:

(7)
\begin{align} \quad y = \frac{1}{2} - x = \frac{1}{2} - \frac{3}{10} = \frac{1}{5} \end{align}

Thus our point of interest is $\left ( \frac{3}{10}, \frac{1}{5} \right )$ just as before, and the minimum value $f$ attains constrained to $g$ is $1$.