Lagrange Multipliers with One Constraint Examples 3

Lagrange Multipliers with One Constraint Examples 3

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 $f(x, y, z) = x^4 + y^4 + z^4$ subject to the constraint $x^2 + y^2 + z^2 = 1$.

Let $g(x, y, z) = x^2 + y^2 + z^2 = 1$. Then in computing the necessary partial derivatives we get that:

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

Assume that $x = 0$ and $y \neq 0$, $z \neq 0$. Then equation 2 implies that $\lambda = 2y^2$ and equation 3 implies that $\lambda = 2z^2$, so $2y^2 = 2z^2$ so $y^2 = z^2$. Plugging this into equation 4 and we have that:

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

Thus the points $\left (0, \pm \frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}} \right )$ are of interest. If we assume that $y = 0$ and $x \neq 0$, $z \neq 0$, as well as assuming that $z = 0$ and $x \neq 0$ with $y \neq 0$, then we have that $\left ( \pm \frac{1}{\sqrt{2}}, 0 , \pm \frac{1}{\sqrt{2}} \right )$ and $\left ( \pm \frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}}, 0 \right )$ are also points of interest.

Now assume that $x = 0$ and $y = 0$. Then $z^2 = \pm 1$, and we have that $(0, 0, \pm 1)$ are points of interest. If we assume that $x = 0$ and $z = 0$ then $y^2 = \pm 1$ and we have that $(0, \pm 1, 0)$ are points of interest. Lastly, if we assume that $y=0$ and $z = 0$ then $x^2 = \pm 1$ and we have that $(\pm 1, 0, 0)$ are points of interest.

Note that we cannot assume that $x = y = z = 0$, since this would contradict equation 4.

Lastly, assume that none of $x$, $y$, and $z$ are zero. Multiply both sides of the first equation by $yz$, both sides of the second equation by $xz$, and both sides of the third equation by $xy$ to get:

(5)
\begin{align} \quad 4x^3yz = \lambda 2xyz \\ \quad 4xy^3z = \lambda 2xyz \\ \quad 4xyz^3 = \lambda 2xyz \\ \quad x^2 + y^2 + z^2 = 1 \end{align}

Thus from above, we can eliminate the multiplier $\lambda$ and we have that:

(6)
\begin{align} \quad x^3yz = xy^3z \\ \quad xy^3z = xyz^3 \\ \quad x^2 + y^2 + z^2 = 1 \end{align}

Equivalently, we can divide equations 1 and equations 2 by $xyz \neq 0$ (since none of $x$, $y$, or $z$ are zero) to get that:

(7)
\begin{align} \quad x^2 = y^2 \\ \quad y^2 = z^2 \\ \quad x^2 + y^2 + z^2 = 1 \end{align}

Thus $y^2 = x^2$ and $z^2 = x^2$. Plugging these into equation three and we have that:

(8)
\begin{align} \quad x^2 + x^2 + x^2 = 1 \\ \quad 3x^2 = 1 \\ \quad x^2 = \frac{1}{3} \\ \quad x = \pm \frac{1}{\sqrt{3}} \end{align}

Thus we have that $\left ( \pm \frac{1}{\sqrt{3}} , \pm \frac{1}{\sqrt{3}} , \pm \frac{1}{\sqrt{3}} \right )$ are points of interest.

As you can see, we have a lot of points of interest. In comparing $f$ evaluated at all of these, we obtain a maximum value of $1$ and a minimum value of $\frac{1}{3}$.