Analytically Determining Extreme Values for Functions of Several Variables

# Analytically Determining Extreme Values for Functions of Several Variables

We will now look at an analytical method for determining if a critical point $(a, b)$ of $z = f(x, y)$ produces a local maximum or local minimum value, or a saddle point. In doing so, consider the difference:

(1)
\begin{align} \quad \Delta f = f(a + h, b + k) - f(a, b) \end{align}
• If $\Delta f > 0$ for small values of $h$ and $k$ then $f$ will have a local minimum at $(a, b)$.
• If $\Delta f < 0$ for small values of $h$ and $k$ then $f$ will have a local maximum at $(a, b)$.
• If $\Delta f > 0$ for some points $(h, k)$ near $(a,b)$ and $\Delta f < 0$ for some points $(h, k)$ near $(a, b)$, then $f$ will have a saddle point at $(a, b)$.

## Example 1

Show that the critical points of the function $f(x, y) = x^3 + y^3 - 3xy$ are $(0, 0)$ and $(1, 1)$ and analytically show that $(0, 0)$ is a saddle point of $f$.

We must first determine the critical points of $f$. To do so, let's find the gradient of $f$ and then set it equal to the zero vector. We have that:

(2)
\begin{align} \quad \nabla f(x, y) = \left ( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right ) = \left ( 3x^2 - 3y, 3y^2 - 3x \right ) \end{align}

Setting the gradient of $f$ equal to the zero vector and we have that:

(3)
\begin{align} \quad \left ( 3x^2 - 3y, 3y^2 - 3x \right ) = (0, 0) \end{align}

We thus obtain the following system of equations:

(4)
\begin{align} \quad 3x^2 - 3y = 0 \implies y = x^2 \\ \quad 3y^2 - 3x = 0 \implies x = y^2 \end{align}

From the equation above we can plug in $y = x^2$ to $x = y^2$ to get:

(5)
\begin{align} \quad x = (x^2)^2 \\ \quad x = x^4 \\ \quad 0 = x^4 - x \\ \quad 0 = x(x^3 - 1) \end{align}

Thus $x = 0$ or $x = 1$. These values of $x$ correspond to the critical points $(0, 0)$ and $(1, 1)$. We are now ready to analytically determine whether these points are local extrema or not.

First consider the critical point $(0, 0)$ and the difference $\Delta f = f(0 + h, 0 + k) - f(0,0)$:

(6)
\begin{align} \quad \Delta f = f(h, k) - f(0, 0) \\ \quad \Delta f = h^3 + k^3 - 3hk - 0 \\ \quad \Delta f = h^3 - 3hk + k^3 \end{align}

Note that $f(h, 0) = h^3 > 0$ for small positive $h$ and $f(h, 0) = h^3 < 0$ for small negative $h$. Thus the point $(0, 0)$ is a saddle point of $f$.