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

Short Answer

Expert verified
The function has a saddle point at \((0, 0)\).

Step by step solution

01

Find the First-Order Partial Derivatives

To find the critical points, we first need to calculate the partial derivatives of the function with respect to each variable. For the function \[ f(x, y) = x^2 + 2xy \]The partial derivative with respect to \( x \) is:\[ f_x = \frac{\partial}{\partial x}(x^2 + 2xy) = 2x + 2y \]And the partial derivative with respect to \( y \) is:\[ f_y = \frac{\partial}{\partial y}(x^2 + 2xy) = 2x \]
02

Solve for Critical Points

Set the first-order partial derivatives equal to zero to find the critical points:1. \( 2x + 2y = 0 \)2. \( 2x = 0 \)From equation 2, we have \( x = 0 \). Substitute \( x = 0 \) into equation 1:\[ 2(0) + 2y = 0 \Rightarrow y = 0 \]Thus, the only critical point is \((0, 0)\).
03

Compute the Second-Order Partial Derivatives

To determine the nature of the critical point \((0,0)\), we need the second-order partial derivatives:\[ f_{xx} = \frac{\partial^2}{\partial x^2}(x^2 + 2xy) = 2 \]\[ f_{yy} = \frac{\partial^2}{\partial y^2}(x^2 + 2xy) = 0 \]\[ f_{xy} = \frac{\partial^2}{\partial x \partial y}(x^2 + 2xy) = 2 \]
04

Apply the Second Derivative Test

The second derivative test formula for functions of two variables is:\[ D = f_{xx}(x_0, y_0)f_{yy}(x_0, y_0) - (f_{xy}(x_0, y_0))^2 \]For the point \((0,0)\):\[ D = (2)(0) - (2)^2 = -4 \]Since \( D < 0 \), \((0, 0)\) is neither a local maximum nor a local minimum, but a saddle point.

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Key Concepts

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

Critical Points
In multivariable calculus, critical points are where the function's rate of change is zero in all directions. To find these points, you must look at the first-order partial derivatives, which indicate the slope of the function relative to each variable. For a function of two variables like \( f(x, y) = x^2 + 2xy \), critical points are where both partial derivatives are zero:
  • \( f_x = 2x + 2y = 0 \)
  • \( f_y = 2x = 0 \)
Solve these equations simultaneously. Here, setting \( 2x = 0 \) gives \( x = 0 \). Substitute \( x = 0 \) into \( 2x + 2y = 0 \) to find \( y = 0 \). Thus, the critical point of the function is \((0, 0)\). Critical points are essential for determining where potential local maxima, minima, or saddle points exist.
Partial Derivatives
Partial derivatives are foundational in multivariable calculus and essential for analyzing functions with more than one variable. They provide information about how a function changes as you change one of its variables while keeping others constant. For the function \( f(x, y) = x^2 + 2xy \), you find the partial derivatives as follows:
  • The partial derivative with respect to \( x \) is \( f_x = 2x + 2y \).
  • The partial derivative with respect to \( y \) is \( f_y = 2x \).
You control one variable at a time and see how it impacts the entire function. This is akin to examining the gradient in multidimensional space, meaning how the function rises or falls as each variable varies.
Second Derivative Test
The second derivative test helps classify the nature of critical points—identifying whether they are maxima, minima, or saddle points. For a function \( f(x, y) \), the second derivative test uses second-order partial derivatives:
  • \( f_{xx} \): Second partial derivative with respect to \( x \).
  • \( f_{yy} \): Second partial derivative with respect to \( y \).
  • \( f_{xy} \): Mixed partial derivative.
The determinant \( D = f_{xx} f_{yy} - (f_{xy})^2 \) is calculated at the critical point. If \( D > 0 \) and \( f_{xx} > 0 \), you have a local minimum. If \( D > 0 \) and \( f_{xx} < 0 \), it's a local maximum. A negative \( D \) indicates a saddle point, where the surface curves up in one direction and down in another. In our example, at \((0, 0)\), \( D = -4 \), leading to a saddle point. Understanding the test allows you to predict and categorize the behavior of multivariable functions effectively.

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