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.) $f(x, y) = 9 - 2x + 4y - x^2 - 4y^2$

• 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 1
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Step-by-step solution:

Step 1: Partial derivatives

To find critical points, calculate the partial derivatives of $f(x, y)$ with respect to $x$ and $y$.

1. Partial derivative with respect to $x$: $f_x = \frac{\partial f}{\partial x} = -2 - 2x$

2. Partial derivative with respect to $y$: $f_y = \frac{\partial f}{\partial y} = 4 - 8y$

Step 2: Solve for critical points

Set $f_x = 0$ and $f_y = 0$ to find the critical points.

1. From $f_x = -2 - 2x = 0$, solve for $x$: $x = -1$

2. From $f_y = 4 - 8y = 0$, solve for $y$: $y = \frac{1}{2}$

Thus, the critical point is: $(x, y) = (-1, \frac{1}{2})$

Step 3: Second partial derivatives

Compute the second partial derivatives to classify the critical point.

1. $f_{xx} = \frac{\partial^2 f}{\partial x^2} = -2$
2. $f_{yy} = \frac{\partial^2 f}{\partial y^2} = -8$
3. $f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0$

Step 4: Discriminant test

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

Substitute the values: $D = (-2)(-8) - (0)^2 = 16$

Since $D > 0$ and $f_{xx} < 0$, the critical point $(-1, \frac{1}{2})$ is a local maximum.

Step 5: Local maximum, minimum, and saddle points

• Local maximum value(s): Substitute $(-1, \frac{1}{2})$ into $f(x, y)$: $f(-1, \frac{1}{2}) = 9 - 2(-1) + 4(\frac{1}{2}) - (-1)^2 - 4(\frac{1}{2})^2$ $= 9 + 2 + 2 - 1 - 1 = 11$

  Therefore, the local maximum value is: $11$

• Local minimum value(s): There are no local minima since $f_{xx} < 0$ at the critical point, indicating a local maximum.

• Saddle point(s): There are no saddle points because $D > 0$ and the critical point is classified as a local maximum.

Final Answer:

Local maximum value(s): $\boxed{11}$

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

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

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