Find the local maximum and minimum values and saddle point(s) of the function. You are encouraged to use a calculator or computer to graph the function with a domain and viewpoint that reveals all the important aspects of the function. (Enter your answers as comma-separated lists. If an answer does not exist, enter DNE.) $f(x, y) = x^4 + y^4 - 4xy + 6$.

• Local maximum value(s):
• Local minimum value(s):
• Saddle point(s) $(x, y) =$

Neetesh Kumar | November 30, 2024

Calculus Homework Help

This is the solution to Math 1D
Assignment: 14.7 Question Number 2
Contact me if you need help with Homework, Assignments, Tutoring Sessions, or Exams for STEM subjects.
You can see our Testimonials or Vouches from here of the previous works I have done.

Get Homework Help

---

Step-by-step solution:

To analyze the given function $f(x, y) = x^4 + y^4 - 4xy + 6$, we follow these steps:

Step 1: Compute the first partial derivatives

The critical points occur where both partial derivatives $f_x$ and $f_y$ are zero.

First partial derivative with respect to $x$:

$f_x = \frac{\partial f}{\partial x} = 4x^3 - 4y$.

First partial derivative with respect to $y$:

$f_y = \frac{\partial f}{\partial y} = 4y^3 - 4x$.

Step 2: Solve for critical points

Set $f_x = 0$ and $f_y = 0$:

1. From $f_x = 4x^3 - 4y = 0$, we get:
   $x^3 = y$.

2. Substitute $y = x^3$ into $f_y = 4y^3 - 4x = 0$:
   $4(x^3)^3 - 4x = 0$,
   $4x^9 - 4x = 0$.

Factor out $4x$:
$4x(x^8 - 1) = 0$.

Solve for $x$:
$x = 0$ or $x^8 = 1$.

For $x^8 = 1$, the solutions are:
$x = \pm 1$.

Thus, the critical points are:
$x = 0, \quad x = 1, \quad x = -1$.

Using $y = x^3$, the corresponding $y$-values are:
For $x = 0: y = 0$,
For $x = 1: y = 1$,
For $x = -1: y = -1$.

The critical points are:
$(0, 0), \quad (1, 1), \quad (-1, -1)$.

Step 3: Compute the second partial derivatives

To classify the critical points, compute the second partial derivatives:
$f_{xx} = \frac{\partial^2 f}{\partial x^2} = 12x^2, \quad f_{yy} = \frac{\partial^2 f}{\partial y^2} = 12y^2, \quad f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -4$.

Step 4: Use the second derivative test

The discriminant $D$ is given by:
$D = f_{xx}f_{yy} - (f_{xy})^2$.

At $(0, 0)$:

$f_{xx} = 0, \quad f_{yy} = 0, \quad f_{xy} = -4$.
$D = (0)(0) - (-4)^2 = -16$.

Since $D < 0$, $(0, 0)$ is a saddle point.

At $(1, 1)$:

$f_{xx} = 12(1)^2 = 12, \quad f_{yy} = 12(1)^2 = 12, \quad f_{xy} = -4$.
$D = (12)(12) - (-4)^2 = 144 - 16 = 128 > 0$.

Since $D > 0$ and $f_{xx} > 0$, $(1, 1)$ is a local minimum.

At $(-1, -1)$:

$f_{xx} = 12(-1)^2 = 12, \quad f_{yy} = 12(-1)^2 = 12, \quad f_{xy} = -4$.
$D = (12)(12) - (-4)^2 = 144 - 16 = 128 > 0$.

Since $D > 0$ and $f_{xx} > 0$, $(-1, -1)$ is a local minimum.

Step 5: Local maximum, minimum, and saddle points

• Local maximum value(s): DNE

• Local minimum value(s): $f(1, 1) = 1^4 + 1^4 - 4(1)(1) + 6 = 1 + 1 - 4 + 6 = 4$,
  $f(-1, -1) = (-1)^4 + (-1)^4 - 4(-1)(-1) + 6 = 1 + 1 - 4 + 6 = 4$.

  Local minima occur at $(1, 1)$ and $(-1, -1)$ with a value of $4$.

• Saddle point(s): $(0, 0)$.

Final Answer:

Local maximum value(s): $\boxed{\text{DNE}}$

Local minimum value(s): $\boxed{4}$

Saddle point(s) $(x, y) =$ $\boxed{(0, 0)}$

Please comment below if you find any error in this solution.
If this solution helps, then please share this with your friends.
Please subscribe to my Youtube channel for video solutions to similar questions.
Keep Smiling :-)