# The Second Derivatives Test for Functions of Two Variables Examples 1

Recall from The Second Derivatives Test for Functions of Two Variables page that if $z = f(x, y)$ is a two variable real-valued function then whose second partial derivatives are continuous on some disk $\mathcal D$ centered at $(a, b)$ where $(a, b)$ is a critical point of $f$ (that is $\nabla f(a, b) = (0, 0)$) and if $D$ is defined to be the function:

(1)

Then we can apply the following second derivatives test:

- If $D(a, b) > 0$ and $\frac{\partial^2}{\partial x^2} f(a, b) > 0$ then $f(a, b)$ is a local minimum value.

- If $D(a, b) > 0$ and $\frac{\partial^2}{\partial y^2} f(a, b) then $f(a, b)$ is a local maximum value.

- If $D(a, b) then $f(a, b)$ is a saddle point.

- If $D(a, b) = 0$ then this test is inconclusive.

We will now look at some examples of finding local maximum and local minimum values of functions of two variables.

## Example 1

**Find any local maximum and local minimum values of the function $f(x, y) = 12x^2 + y^3 -12xy$.**

We should first note that this function is continuous and defined for all $(x, y) \in \mathbb{R}^2$ (since $f$ is a polynomial) and so we do not need to worry about singular points or boundary points. Thus, any local extrema will occur at critical points. Let’s find these critical points by finding the points $(x, y)$ for which the gradient $\nabla f(x, y) = (0, 0)$. We first find the partial derivatives of $f$ to get that:

(2)

Setting the gradient $\nabla f(x, y) = (0, 0)$ we get that $24x – 12y = 0$ and $3y^2 – 12x = 0$. The first equation gives us that $y = 2x$, and plugging this into the second equation gives us the polynomial $3(2x)^2 – 12x = 12x^2 – 12x = 12x(x – 1) = 0$. Thus $x = 0$ and $x = 1$. Therefore our critical points are $(0, 0)$ and $(1, 2)$.

Now let’s determine the function $D(x, y)$. To do so, we need to first compute the second partial derivatives of $f$:

(3)

Therefore we have that:

(4)

We see that $D(0, 0) = -144 and so $(0, 0)$ is a saddle point of $f$. We also have that $D(1, 2) = 144 > 0$ and $\frac{\partial^2}{\partial x^2} f(1, 2) = 24 > 0$ and so $(1, 2)$ produces a local minimum value, in particular, $f(1, 2) = -4$ is the local minimum value.

## Example 2

**Find any local maximum and local minimum values of the function $f(x, y) = 2x^2 – 2x^2y + y^2$.**

Once again, this function is continuous and defined for all $(x, y) \in \mathbb{R}^2$ so we do not need to worry about singular points or boundary points. Let’s find the critical points of $f$ by setting the gradient of $f$ equal to zero.

(5)

Thus we obtain the equations $4x – 4xy = 0$ and $-2x^2 +2y = 0$. The second equation says that $y = x^2$, and substituting this into the first equation and we get that $4x – 4x(x^2) = 4x – 4x^3 = 4x(1 – x^2) = 0$. Thus we have that $x = 0$, $x = 1$, and $x = -1$. So we have three critical points which are $(0, 0)$, $(1, 1)$, and $(-1, 1)$.

Now we will compute the second partial derivatives:

(6)

Therefore $D(x, y)$ is as follows:

(7)

Hence we see that $D(0, 0) = 8 > 0$ and $\frac{\partial^2 }{\partial x^2} f(0, 0) = 4 > 0$ and thus $f(0, 0) = 0$ is a local minimum value of $f$. We also have that $D(1, 1) = -16 and so $f(1, 1) = 1$ is a saddle point of $f$. Lastly, $D(-1, 1) = -16 , so $f(1, -1) = 1$ is also a saddle point of $f$