Question

Find all points (x, y) where f(x, y) has a possible relative maximum or minimum.

f(x, y)=1/3 x³-5 y³-2 x+15 y-6
What are all the possible point(s)?

Answer 1

We have four critical points:

1. (x, y) = (√2, 1)

2. (x, y) = (√2, -1)

3. (x, y) = (-√2, 1)

4. (x, y) = (-√2, -1)

Step-by-step explanation:

To find the possible relative maximum or minimum points of the function f(x, y) = 1/3x³ - 5y³ - 2x + 15y - 6, we need to find the critical points where the partial derivatives are equal to zero or do not exist.

First, we find the partial derivative with respect to x:

∂f/∂x = x² - 2

Then, we find the partial derivative with respect to y:

∂f/∂y = -15y² + 15

To find the critical points, we set both partial derivatives equal to zero and solve for x and y separately.

For ∂f/∂x = x² - 2 = 0:

x² = 2

x = ±√2

For ∂f/∂y = -15y² + 15 = 0:

-15y² = -15

y² = 1

y = ±1

Therefore, we have four critical points:

1. (x, y) = (√2, 1)

2. (x, y) = (√2, -1)

3. (x, y) = (-√2, 1)

4. (x, y) = (-√2, -1)

((To determine if these points are relative maximum or minimum points, we need to further analyze the function using the second partial derivatives test or check the behavior of the function around these points. However, since the question asks for all possible points, the four critical points mentioned above are the only points to be considered.))

Answer 2

The function
`f(x, y) = (1/3)x³ - 5y³ - 2x + 15y - 6` has possible relative maximum or minimum at the points
`(√2, 1), (-√2, 1), (√2, -1)`,

and
`(-√2, -1).`

Step-by-step explanation:

To find the points where the function
`f(x, y) = (1/3)x³ - 5y³ - 2x + 15y - 6` has a possible relative maximum or minimum, we need to find the critical points. The critical points occur where the partial derivatives of the function are equal to zero.

Let's find the partial derivatives:

`fx(x, y) = x² - 2`

`fy(x, y) = -15y² + 15`

We set both partial derivatives equal to zero and solve for
`x` and
`y`:

`x² - 2 = 0`

`-15y² + 15 = 0`

Solving these equations, we get:

`x = ±√2y = ±1`

So, the possible points where
`f(x, y)` has a relative maximum or minimum are:
`(√2, 1)(-√2, 1)(√2, -1)(-√2, -1)`