Math / Calculus

Question3. [-/2 Points] DETAILS MY NOTES MARSVECTORCALC6 3.3.004. PRACTICE ANOTHER
Find the critical point of the given function and then determine whether it is a local maximum, local minimum, or saddle point. $f(x, y)=x^{2}+y^{2}+8 x y$ \begin{tabular}{|c|c|} \hline critical point & classification \\ \hline $(x, y)=(\square)$ & Select Classification $\widehat{ }$ \\ \hline \end{tabular}
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Studdy Solution

STEP 1

What is this asking? Find the mysterious point where this 3D curvy surface $f(x,y)$ flattens out, and then tell us if it's a peak, a valley, or a mountain pass! Watch out! Don't mix up the tests for local max/min versus saddle points.
Also, make sure your partial derivatives are on point!

STEP 2

1. Find the critical points
2. Classify the critical point

STEP 3

Let's kick things off by finding the partial derivative of $f(x,y)$ with respect to $x$ .
Imagine $y$ is just chilling as a constant, and differentiate with respect to $x$ . $\frac{\partial f}{\partial x} = 2x + 8y$

STEP 4

Now, let's find the partial derivative of $f(x,y)$ with respect to $y$ .
This time, $x$ is hanging out as a constant while we differentiate with respect to $y$ . $\frac{\partial f}{\partial y} = 2y + 8x$

STEP 5

To find the critical points, we need to set both partial derivatives equal to zero and solve the resulting system of equations.
This is where the magic happens! $2x + 8y = 0$ $2y + 8x = 0$

STEP 6

From the first equation, we can express $x$ in terms of $y$ : $2x = -8y$ , so $x = -4y$ .
Now, substitute this into the second equation: $2y + 8(-4y) = 0$ $2y - 32y = 0$ $-30y = 0$ So, $y = 0$ !
Since $x = -4y$ , we also have $x = -4 \cdot 0 = 0$ .
Therefore, our critical point is $(0, 0)$ !

STEP 7

To classify the critical point, we need the second partial derivatives.
Buckle up! $\frac{\partial^2 f}{\partial x^2} = 2$ $\frac{\partial^2 f}{\partial y^2} = 2$ $\frac{\partial^2 f}{\partial x \partial y} = 8$

STEP 8

Now, let's calculate the discriminant, often denoted as $D$ .
It's like a secret code to unlock the nature of our critical point! $D = \left(\frac{\partial^2 f}{\partial x^2}\right)\left(\frac{\partial^2 f}{\partial y^2}\right) - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2$ $D = (2)(2) - (8)^2$ $D = 4 - 64$ $D = -60$

STEP 9

Since $D < 0$ , our critical point $(0, 0)$ is a saddle point!

STEP 10

Critical point: $(x, y) = (0, 0)$ .
Classification: Saddle Point.

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Studdy Solution

STEP 1

What is this asking? Find the mysterious point where this 3D curvy surface $f(x,y)$ flattens out, and then tell us if it's a peak, a valley, or a mountain pass! Watch out! Don't mix up the tests for local max/min versus saddle points.
Also, make sure your partial derivatives are on point!

STEP 2

1. Find the critical points
2. Classify the critical point

STEP 3

Let's kick things off by finding the partial derivative of $f(x,y)$ with respect to $x$ .
Imagine $y$ is just chilling as a constant, and differentiate with respect to $x$ . $\frac{\partial f}{\partial x} = 2x + 8y$

STEP 4

Now, let's find the partial derivative of $f(x,y)$ with respect to $y$ .
This time, $x$ is hanging out as a constant while we differentiate with respect to $y$ . $\frac{\partial f}{\partial y} = 2y + 8x$

STEP 5

To find the critical points, we need to set both partial derivatives equal to zero and solve the resulting system of equations.
This is where the magic happens! $2x + 8y = 0$ $2y + 8x = 0$

STEP 6

From the first equation, we can express $x$ in terms of $y$ : $2x = -8y$ , so $x = -4y$ .
Now, substitute this into the second equation: $2y + 8(-4y) = 0$ $2y - 32y = 0$ $-30y = 0$ So, $y = 0$ !
Since $x = -4y$ , we also have $x = -4 \cdot 0 = 0$ .
Therefore, our critical point is $(0, 0)$ !

STEP 7

To classify the critical point, we need the second partial derivatives.
Buckle up! $\frac{\partial^2 f}{\partial x^2} = 2$ $\frac{\partial^2 f}{\partial y^2} = 2$ $\frac{\partial^2 f}{\partial x \partial y} = 8$

STEP 8

STEP 9

Since $D < 0$ , our critical point $(0, 0)$ is a saddle point!

STEP 10

Critical point: $(x, y) = (0, 0)$ .
Classification: Saddle Point.

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Studdy Solution

STEP 1

What is this asking? Find the mysterious point where this 3D curvy surface f(x,y)f(x,y)f(x,y) flattens out, and then tell us if it's a peak, a valley, or a mountain pass! Watch out! Don't mix up the tests for local max/min versus saddle points.Also, make sure your partial derivatives are on point!

STEP 2

1. Find the critical points2. Classify the critical point

STEP 3

Let's **kick things off** by finding the partial derivative of f(x,y)f(x,y)f(x,y) with respect to xxx.Imagine yyy is just chilling as a constant, and differentiate with respect to xxx. ∂f∂x=2x+8y \frac{\partial f}{\partial x} = 2x + 8y ∂x∂f​=2x+8y

STEP 4

Now, let's find the partial derivative of f(x,y)f(x,y)f(x,y) with respect to yyy.This time, xxx is hanging out as a constant while we differentiate with respect to yyy. ∂f∂y=2y+8x \frac{\partial f}{\partial y} = 2y + 8x ∂y∂f​=2y+8x

STEP 5

To **find the critical points**, we need to set both partial derivatives equal to zero and solve the resulting system of equations.This is where the magic happens! 2x+8y=0 2x + 8y = 0 2x+8y=0 2y+8x=0 2y + 8x = 0 2y+8x=0

STEP 6

STEP 7

STEP 8

STEP 9

Since D<0D < 0D<0, our critical point (0,0)(0, 0)(0,0) is a **saddle point**!

STEP 10

Critical point: (x,y)=(0,0)(x, y) = (0, 0)(x,y)=(0,0).Classification: Saddle Point.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? Find the mysterious point where this 3D curvy surface $f(x,y)$ flattens out, and then tell us if it's a peak, a valley, or a mountain pass! Watch out! Don't mix up the tests for local max/min versus saddle points.
Also, make sure your partial derivatives are on point!

1. Find the critical points
2. Classify the critical point

Let's kick things off by finding the partial derivative of $f(x,y)$ with respect to $x$ .
Imagine $y$ is just chilling as a constant, and differentiate with respect to $x$ . $\frac{\partial f}{\partial x} = 2x + 8y$

Now, let's find the partial derivative of $f(x,y)$ with respect to $y$ .
This time, $x$ is hanging out as a constant while we differentiate with respect to $y$ . $\frac{\partial f}{\partial y} = 2y + 8x$

To find the critical points, we need to set both partial derivatives equal to zero and solve the resulting system of equations.
This is where the magic happens! $2x + 8y = 0$ $2y + 8x = 0$

Since $D < 0$ , our critical point $(0, 0)$ is a saddle point!

Critical point: $(x, y) = (0, 0)$ .
Classification: Saddle Point.