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

Short Answer

Expert verified
Local minimum at \((0,0)\), saddle point at \((0,2)\).

Step by step solution

01

Calculate First Derivatives

To find the critical points, we first need to compute the partial derivatives of the function. The function is \( f(x, y) = e^{-y}(x^2 + y^2) \). Compute \( f_x \), the partial derivative with respect to \( x \), and \( f_y \), the partial derivative with respect to \( y \). \( f_x = \frac{\partial}{\partial x} e^{-y}(x^2 + y^2) = e^{-y} \cdot (2x) = 2xe^{-y} \)\( f_y = \frac{\partial}{\partial y} e^{-y}(x^2 + y^2) = e^{-y} (0 + 2y) - e^{-y}(x^2 + y^2) = e^{-y} (2y - x^2 - y^2) \)
02

Solve First Derivatives

Set the first partial derivatives equal to zero to find the critical points: \[ 2xe^{-y} = 0 \]\[ e^{-y}(2y - x^2 - y^2) = 0 \].From the first equation, we have:\1. \( 2xe^{-y} = 0 \) implies \( x = 0 \) since \( e^{-y} eq 0 \).Substitute \( x = 0 \) into the second equation:\( 2ye^{-y} = e^{-y} y^2 \)This simplifies to:\( 2y = y^2 \),\( y(y-2) = 0 \).So, \( y = 0 \) or \( y = 2 \).Thus, the critical points are \((0,0)\) and \((0,2)\).
03

Calculate Second Derivatives

To determine the nature of the critical points, compute the second partial derivatives. - \( f_{xx} = \frac{\partial}{\partial x} (2xe^{-y}) = 2e^{-y} \)- \( f_{yy} = \frac{\partial}{\partial y} \left( e^{-y}(2y - x^2 - y^2) \right) = e^{-y}[2 - (2y - x^2 - y^2)] = e^{-y}(-y^2 - x^2 + 2) \)Additionally, compute \( f_{xy} = \frac{\partial}{\partial y} (2xe^{-y}) = -2xe^{-y} \).
04

Use the Second Derivative Test

Use the second derivative test to classify the critical points. Define the Hessian determinant \( H \) as:\[ H = f_{xx}f_{yy} - (f_{xy})^2 \]Evaluate at \((0,0)\):- \( f_{xx}(0,0) = 2 \)- \( f_{yy}(0,0) = 2 \)- \( f_{xy}(0,0) = 0 \)- \( H(0,0) = 2 \times 2 - 0^2 = 4 \), which is positive, and \( f_{xx} > 0 \), so \((0,0)\) is a local minimum.Evaluate at \((0,2)\):- \( f_{xx}(0,2) = 2e^{-2} \)- \( f_{yy}(0,2) = -4e^{-2} \)- \( f_{xy}(0,2) = 0 \)- \( H(0,2) = 2e^{-2} \times (-4e^{-2}) - 0^2 = -8e^{-4} \), which is negative, so \((0,2)\) is a saddle point.

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Key Concepts

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

second derivative test
In multivariable calculus, the second derivative test is a valuable tool for classifying critical points of functions. Once you have found the critical points through the first derivatives, the next step is to determine their nature using the second derivatives.
Let us consider what is known as Hessian determinant. The Hessian matrix is composed of the second-order partial derivatives of the function. For functions of two variables, the Hessian determinant \( H \) is given by:
\[H = f_{xx}f_{yy} - (f_{xy})^2\]where \( f_{xx} \), \( f_{yy} \), and \( f_{xy} \) represent the second partial derivatives of the function, measured at the critical points.
Given:
  • \( H > 0 \) and \( f_{xx} > 0 \): The point is a local minimum.
  • \( H > 0 \) and \( f_{xx} < 0 \): The point is a local maximum.
  • \( H < 0 \): The point is a saddle point, indicating a change in concavity.
  • \( H = 0 \): The test is inconclusive.
With these rules, you can classify critical points effectively and understand the behavior of the function around these points.
partial derivatives
Partial derivatives are fundamental in understanding functions of multiple variables. They provide insights into how the function changes when only one variable is varied, while keeping the others constant.
The notation \( \frac{\partial f}{\partial x} \) represents the partial derivative of \( f \) with respect to \( x \). For the given function \( f(x, y) = e^{-y}(x^2 + y^2) \), the partial derivative with respect to \( x \) is calculated by treating \( y \) as a constant, resulting in:
\[f_x = 2xe^{-y}\]Similarly, the partial derivative with respect to \( y \) is obtained by treating \( x \) as a constant:
\[f_y = e^{-y} (2y - x^2 - y^2)\]These computations are crucial for determining the critical points of a function, as they help identify where the function does not change to the first order - these points are potential candidates for local maxima, minima, or saddle points.
local minima
Understanding local minima is crucial in identifying points where a function attains a minimum value in its immediate vicinity. Local minimum points are characterized by the fact that the function is not necessarily at its lowest value globally, but is lower than neighboring points. In our function \( f(x, y) = e^{-y}(x^2 + y^2) \), the critical point \((0,0)\) is a candidate for being a local minimum.
To verify this, we turn to the second derivative test. For a point to be classified as a local minimum, the Hessian determinant \( H \) evaluated at this point should be positive, and \( f_{xx} \) should also be positive:
  • At \((0, 0)\), \( H = 4 \) (which is positive).
  • The second derivative \( f_{xx} \) at \((0, 0)\) is \( 2 \), indicating a concave up shape in the \( x \)-direction.
These observations confirm that \((0,0)\) is indeed a local minimum. Understanding this concept helps in analyzing the function's behavior at key points and knowing their importance in optimization problems.
saddle points
Saddle points are unique critical points where the function does not exhibit the extremeness of either a local minima or maxima. Instead, they represent a point where the function surface curves up in one direction and down in another.
For the function \( f(x, y) = e^{-y}(x^2 + y^2) \), the critical point \((0,2)\) is an example of a saddle point.
In using the second derivative test to classify this critical point, it is found that:
  • At \((0, 2)\), the Hessian determinant \( H = -8e^{-4} \), which is negative.
A negative \( H \) implies mixed curvature at the point, a defining characteristic of saddle points. This gives a clear indication that the surface goes up in one direction and down in another, much like the shape of a saddle. Thus, understanding saddle points helps in discerning the complex topology of a multivariable function.

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