Critical Points Of F(x,y) = 2x^3 - Y^2 + 6x^2 + 4y + 11

Hey math whizzes, let's dive into the fascinating world of calculus and tackle a problem involving finding and classifying the critical points of a multivariable function. Our mission today is to dissect the function $f(x, y)=2 x^3-y^2+6 x^2+4 y+11$. We'll be hunting for those special points where the function's behavior might change, which we call critical points. Once we find them, we'll put on our detective hats and determine if each point is a local minimum, a local maximum, or a saddle point. So, grab your calculators, sharpen your pencils, and let's get started on this mathematical adventure!

Finding the Critical Points: Where the Action Happens

Alright guys, the first step in our quest to understand the function $f(x, y)=2 x^3-y^2+6 x^2+4 y+11$ is to locate its critical points. What exactly are critical points, you ask? Well, they're the spots on the graph of our function where things get interesting. Mathematically, critical points are the points $(x, y)$ in the domain of $f$ where either the gradient of $f$ is the zero vector (meaning both partial derivatives are zero), or where at least one of the partial derivatives does not exist. For our nicely behaved polynomial function, we don't need to worry about the derivatives not existing, so we're solely focused on finding where the gradient is zero. This means we need to find the points $(x, y)$ where both $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$. Let's get our hands dirty and compute these partial derivatives. The partial derivative of $f$ with respect to $x$, denoted as $f_x$, is found by treating $y$ as a constant and differentiating with respect to $x$. So, $f_x = \frac{\partial}{\partial x}(2 x^3-y^2+6 x^2+4 y+11) = 2(3x^2) - 0 + 6(2x) + 0 + 0 = 6x^2 + 12x$. Now, let's find the partial derivative of $f$ with respect to $y$, denoted as $f_y$. Here, we treat $x$ as a constant and differentiate with respect to $y$. So, $f_y = \frac{\partial}{\partial y}(2 x^3-y^2+6 x^2+4 y+11) = 0 - 2y + 0 + 4 + 0 = -2y + 4$. To find our critical points, we set both of these partial derivatives equal to zero and solve the resulting system of equations:

$6x^2 + 12x = 0$ $-2y + 4 = 0$

Let's tackle the second equation first because it looks much simpler. From $-2y + 4 = 0$, we can easily solve for $y$: $-2y = -4$, which gives us $y = 2$. Now, let's move on to the first equation: $6x^2 + 12x = 0$. We can factor out a $6x$ from both terms: $6x(x + 2) = 0$. This equation holds true if either $6x = 0$ or $x + 2 = 0$. If $6x = 0$, then $x = 0$. If $x + 2 = 0$, then $x = -2$.

So, we have found our possible $x$ values for the critical points: $x = 0$ and $x = -2$. Since we found that $y$ must be $2$ for any critical point, we can now combine these values to get our critical points. When $x = 0$ and $y = 2$, we get the point $(0, 2)$. When $x = -2$ and $y = 2$, we get the point $(-2, 2)$. These are the candidates for our local minima, maxima, or saddle points. It's crucial to remember that finding where the gradient is zero is just the first step; the real fun begins when we classify these points!

Classifying the Critical Points: Minima, Maxima, or Saddle?

Now that we've successfully identified our critical points, $(0, 2)$ and $(-2, 2)$, it's time for the exciting part: classifying them! To do this, we'll employ the Second Derivative Test, which is a super handy tool in multivariable calculus. This test involves calculating the second partial derivatives of our function $f(x, y)$ and then evaluating a special quantity called the discriminant, or Hessian determinant, at each critical point. The discriminant, denoted by $D$, is defined as $D(x, y) = f_{xx}(x, y) f_{yy}(x, y) - [f_{xy}(x, y)]^2$. Here, $f_{xx}$ is the second partial derivative with respect to $x$ twice, $f_{yy}$ is the second partial derivative with respect to $y$ twice, and $f_{xy}$ is the mixed second partial derivative (first with respect to $x$, then $y$, or vice versa - they'll be the same for our polynomial!).

Let's calculate these second partial derivatives. We already found $f_x = 6x^2 + 12x$ and $f_y = -2y + 4$. Now, we find $f_{xx}$ by differentiating $f_x$ with respect to $x$: $f_{xx} = \frac{\partial}{\partial x}(6x^2 + 12x) = 12x + 12$. Next, we find $f_{yy}$ by differentiating $f_y$ with respect to $y$: $f_{yy} = \frac{\partial}{\partial y}(-2y + 4) = -2$. Finally, let's find the mixed partial derivative $f_{xy}$. We differentiate $f_x$ with respect to $y$: $f_{xy} = \frac{\partial}{\partial y}(6x^2 + 12x) = 0$. Alternatively, we could find $f_{yx}$ by differentiating $f_y$ with respect to $x$: $f_{yx} = \frac{\partial}{\partial x}(-2y + 4) = 0$. As expected for a polynomial, $f_{xy} = f_{yx}$.

Now we can compute the discriminant $D(x, y) = f_{xx}(x, y) f_{yy}(x, y) - [f_{xy}(x, y)]^2$. Plugging in our second partial derivatives, we get $D(x, y) = (12x + 12)(-2) - (0)^2 = -2(12x + 12) = -24x - 24$.

The Second Derivative Test rules are as follows:

1. If $D(a, b) > 0$ and $f_{xx}(a, b) > 0$, then $f$ has a local minimum at $(a, b)$.
2. If $D(a, b) > 0$ and $f_{xx}(a, b) < 0$, then $f$ has a local maximum at $(a, b)$.
3. If $D(a, b) < 0$, then $f$ has a saddle point at $(a, b)$.
4. If $D(a, b) = 0$, the test is inconclusive, and we can't determine the nature of the critical point using this method alone.

Let's apply these rules to our critical points!

1. For the critical point $(0, 2)$: First, we evaluate the discriminant $D$ at $(0, 2)$: $D(0, 2) = -24(0) - 24 = -24$. Since $D(0, 2) < 0$, according to rule 3, the critical point $(0, 2)$ is a saddle point. This means that in some directions, the function increases as you move away from this point, and in other directions, it decreases.

2. For the critical point $(-2, 2)$: Next, we evaluate the discriminant $D$ at $(-2, 2)$: $D(-2, 2) = -24(-2) - 24 = 48 - 24 = 24$. Since $D(-2, 2) > 0$, we need to check the sign of $f_{xx}(-2, 2)$ to determine if it's a minimum or maximum. We found $f_{xx}(x, y) = 12x + 12$. So, $f_{xx}(-2, 2) = 12(-2) + 12 = -24 + 12 = -12$. Since $D(-2, 2) > 0$ and $f_{xx}(-2, 2) < 0$, according to rule 2, the critical point $(-2, 2)$ is a local maximum. This point represents a peak on the surface of our function.

Summary of Findings

To wrap things up, guys, we successfully navigated the process of finding and classifying the critical points for the function $f(x, y)=2 x^3-y^2+6 x^2+4 y+11$.

We found the critical points by setting the first partial derivatives $f_x$ and $f_y$ equal to zero. This led us to a system of equations that yielded two critical points: $(0, 2)$ and $(-2, 2)$.

Then, we applied the powerful Second Derivative Test. By calculating the discriminant $D(x, y) = f_{xx}f_{yy} - (f_{xy})^2$, we were able to classify these points:

• At the point $(0, 2)$: We found $D(0, 2) = -24$, which is negative. This clearly indicates that $(0, 2)$ is a saddle point.
• At the point $(-2, 2)$: We found $D(-2, 2) = 24$, which is positive. We also checked $f_{xx}(-2, 2) = -12$, which is negative. Therefore, $(-2, 2)$ is a local maximum.

So there you have it! For the function $f(x, y)=2 x^3-y^2+6 x^2+4 y+11$, the point $(0, 2)$ is a saddle point, and the point $(-2, 2)$ is a local maximum. Understanding these critical points is fundamental to analyzing the behavior of functions in multivariable calculus, giving us insights into where the function reaches its highest or lowest values, or where it exhibits more complex behavior like saddle points. Keep practicing, and you'll master these techniques in no time!