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Find all the local maxima, local minima, and saddle points of the functions. $$ f(x, y)=x^{3}+y^{3}+3 x^{2}-3 y^{2}-8 $$

Short Answer

Expert verified
(0, 2) is a local minimum, (-2, 0) is a local maximum, (0, 0) and (-2, 2) are saddle points.

Step by step solution

01

Find the First Partial Derivatives

First, we need to compute the first partial derivatives of the function with respect to both variables, \(x\) and \(y\). The partial derivative with respect to \(x\) is \(f_x = \frac{\partial f}{\partial x} = 3x^2 + 6x\). The partial derivative with respect to \(y\) is \(f_y = \frac{\partial f}{\partial y} = 3y^2 - 6y\).
02

Set First Partial Derivatives to Zero

To find the critical points, set the partial derivatives equal to zero: \(f_x = 3x^2 + 6x = 0\) and \(f_y = 3y^2 - 6y = 0\). Solving these equations, we find: \(x(3x + 6) = 0\) and \(y(3y - 6) = 0\). This gives the solutions \(x = 0\), \(x = -2\) and \(y = 0\), \(y = 2\).
03

Find Critical Points

Combining the solutions from step 2, we find the critical points: \((0, 0)\), \((0, 2)\), \((-2, 0)\), \((-2, 2)\).
04

Compute the Second Partial Derivatives

Next, compute the second partial derivatives: \(f_{xx} = 6x + 6\), \(f_{yy} = 6y - 6\), and \(f_{xy} = 0\).
05

Apply the Second Derivative Test

For each critical point, use the second derivative test. Compute the Hessian determinant, \(D = f_{xx}f_{yy} - (f_{xy})^2\).For \((0, 0)\): \(f_{xx} = 6\), \(f_{yy} = -6\), \(D = 6(-6) - 0^2 = -36\). This is a saddle point.For \((0, 2)\): \(f_{xx} = 6\), \(f_{yy} = 6\), \(D = 6 \times 6 - 0^2 = 36\) and \(f_{xx} = 6 > 0\). This is a local minimum.For \((-2, 0)\): \(f_{xx} = -6\), \(f_{yy} = -6\), \(D = (-6)(-6) - 0^2 = 36\) and \(f_{xx} = -6 < 0\). This is a local maximum.For \((-2, 2)\): \(f_{xx} = -6\), \(f_{yy} = 6\), \(D = (-6)(6) - 0^2 = -36\). This is a saddle point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Partial Derivatives
When we analyze a function with multiple variables, partial derivatives play a crucial role. They allow us to understand how the function changes as each independent variable changes, while keeping the others constant. The first partial derivatives are the derivatives of a function with respect to each of its variables. In our case, the function is \(f(x, y) = x^3 + y^3 + 3x^2 - 3y^2 - 8\).

To find the first partial derivatives, we compute them separately for \(x\) and \(y\). The partial derivative with respect to \(x\) is given by \(f_x = \frac{\partial f}{\partial x} = 3x^2 + 6x\), and with respect to \(y\) it is \(f_y = \frac{\partial f}{\partial y} = 3y^2 - 6y\).

These derivatives provide us with rates of change along the \(x\) and \(y\) axes. By setting these equations to zero, we determine the critical points, which are potential candidates for maxima, minima, or saddle points. These critical points are essential for further analysis to understand the behavior of the function at certain locations.
Second Derivative Test
After finding critical points using the first partial derivatives, the second derivative test helps us determine their nature. This test uses second partial derivatives to classify critical points as local maxima, local minima, or saddle points.

We compute the second partial derivatives. For our function \(f(x, y)\), these are:
  • \(f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x + 6\)
  • \(f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y - 6\)
  • \(f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0\)
By analyzing these values at each critical point, we can find out if it's a maximum, minimum, or saddle point by considering the second derivative test. This involves comparing the signs and magnitudes of these derivatives.
Hessian Determinant
The Hessian determinant is a powerful tool in multivariable calculus used for determining the behavior of critical points. It is denoted as \(D\) and is calculated using the second partial derivatives of a function. For a two-variable function, as in our example, the Hessian determinant is calculated by:

\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]

The sign and value of \(D\) help determine the type of critical point. Here's how:
  • If \(D > 0\) and \(f_{xx} > 0\), the function has a local minimum at the critical point.
  • If \(D > 0\) and \(f_{xx} < 0\), the function has a local maximum.
  • If \(D < 0\), the function has a saddle point.
  • If \(D = 0\), the test is inconclusive.
Using the Hessian determinant gives us a systematic way to classify critical points, which is crucial in understanding the behavior and shape of the function at these locations.

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