Question

Determine the stationary values for z = x³ - 3x + xy² and state their nature (i.e., maximum, minimum, saddle point).

Answer 1

To determine the stationary values of z = x³ - 3x + xy², we need to find the critical points by setting the partial derivatives equal to zero. After finding the critical points, we can classify their nature by using the second derivative test.

Step-by-step explanation:

To determine the stationary values of z = x³ - 3x + xy², we need to find the critical points by setting the partial derivatives equal to zero. Taking the derivative with respect to x, we get dz/dx = 3x² - 3 + y², and the derivative with respect to y gives dz/dy = 2xy. Setting both derivatives equal to zero and solving the system of equations will give us the critical points. After finding the critical points, we can classify their nature by using the second derivative test.

If we solve the system of equations dz/dx = 3x² - 3 + y² = 0 and dz/dy = 2xy = 0, we find two critical points: (1, 0) and (-1, 0). To classify their nature, we need to evaluate the second derivatives: d²z/dx² = 6x and d²z/dy² = 2x. Plug in the values of x from the critical points into the second derivatives. At (1, 0), d²z/dx² = 6(1) = 6, which is positive, indicating a minimum. At (-1, 0), d²z/dx² = 6(-1) = -6, which is negative, indicating a maximum.

Answer 2

The stationary points for
`\( z = x^3 - 3x + xy^2 \)` are at
`\((1, -1)\)` and
`\((1, 1)\)`, and their nature is as follows:

-
`\( (1, -1) \)` is a saddle point.

-
`\( (1, 1) \)` is a saddle point.

Step-by-step explanation:

To find the stationary points, partial differentiation is used to get the critical points where both partial derivatives
`\( (\partial z)/(\partial x) \)`and \( \frac{\partial z}
`{\partial y} \)` are equal to zero.

Given
`\( z = x^3 - 3x + xy^2 \)`, compute the partial derivatives:

`\[ (\partial z)/(\partial x) = 3x^2 - 3 + y^2 \]`

`\[ (\partial z)/(\partial y) = 2xy \]`

Setting these derivatives equal to zero and solving for x and y:

`\[ 3x^2 - 3 + y^2 = 0 \]`

`\[ 2xy = 0 \]`

From the equation
`\(2xy = 0\)`, either x = 0 or y = 0. However, when y = 0, it doesn't satisfy
`\(3x^2 - 3 + y^2 = 0\)`. Hence, y ≠ 0.

When
`\(3x^2 - 3 + y^2 = 0\)` is rearranged,
`\(3x^2 + y^2 = 3\)`, which implies
`\(x^2 + (y^2)/(3) = 1\)`. This equation represents an ellipse centered at the origin with major axis along the x-axis.

Solving for y in terms of x,
`\(y = \pm √(3 - 3x^2)\)`.

For
`\(y = √(3 - 3x^2)\)`, at x = 1, y = 0. Hence, one point is (1, 0).

For
`\(y = -√(3 - 3x^2)\)`, at x = 1, y = 0. Hence, the other point is also (1, 0).

Substituting these points into
`\(z = x^3 - 3x + xy^2\)`:

At (1, 0),
`\(z = 1^3 - 3(1) + 1(0)^2 = -2\)`.

Therefore, the stationary point is (1, 0), and it's a saddle point.

Hence, the stationary points for
`\( z = x^3 - 3x + xy^2 \)` are at
`\((1, -1)\)` and
`\((1, 1)\),` and both are saddle points.