Problem 25 - Calculus and its applications

Question

Find the relative maximum and minimum values as well as any saddle points. $$ f(x, y)=2 y^{2}+x^{2}-x^{2} y $$

Step-by-Step Solution

Answer

The function has a relative minimum at \((0, 0)\) and a saddle point at \((2, 1)\).

Step 1: Find the First Partial Derivatives

First, we need to find the first partial derivatives of the function \( f(x, y) = 2y^2 + x^2 - x^2y \). The partial derivative with respect to \( x \) is \( f_x(x, y) = 2x - 2xy \), and with respect to \( y \) is \( f_y(x, y) = 4y - x^2 \).

Step 2: Set First Partial Derivatives to Zero

To find the critical points, we set the first partial derivatives to zero:1. \( 2x - 2xy = 0 \)2. \( 4y - x^2 = 0 \).Solving \( f_x(x, y) = 0 \) implies \( 2x(1 - y) = 0 \), which gives us \( x = 0 \) or \( y = 1 \). Then, solving \( f_y(x, y) = 0 \) with these cases gives us points \((0, 0)\) and \((2, 1)\).

Step 3: Find the Second Partial Derivatives

Next, find the second partial derivatives to set up the Hessian:- \( f_{xx} = \frac{\partial}{\partial x}(2x - 2xy) = 2 - 2y \).- \( f_{yy} = \frac{\partial}{\partial y}(4y - x^2) = 4 \).- \( f_{xy} = \frac{\partial}{\partial y}(2x - 2xy) = -2x \).

Step 4: Calculate the Hessian Determinant

The Hessian determinant is given by \( H = f_{xx}f_{yy} - (f_{xy})^2 \):- For \((0, 0)\): - \( f_{xx} = 2, f_{yy} = 4, f_{xy} = 0 \). - Hessian: \( H = (2)(4) - (0)^2 = 8 \).- For \((2, 1)\): - \( f_{xx} = 0, f_{yy} = 4, f_{xy} = -4 \). - Hessian: \( H = (0)(4) - (-4)^2 = -16 \).

Step 5: Classify Critical Points

Use the Hessian to classify each critical point:- \((0, 0)\) has a positive Hessian and \( f_{xx} > 0 \), indicating a local minimum.- \((2, 1)\) has a negative Hessian, indicating a saddle point.

Key Concepts

Partial Derivatives

Partial derivatives are essential tools in calculus, especially when working with functions of multiple variables. In this case, the function is given as \( f(x, y) = 2y^2 + x^2 - x^2y \). Partial derivatives help us explore how the function changes when only one variable is adjusted while keeping the other constant.

To find the partial derivative with respect to \( x \), treat \( y \) as a constant and differentiate the function:

• \( f_x(x, y) = \frac{\partial}{\partial x}(2y^2 + x^2 - x^2y) = 2x - 2xy \)

For the partial derivative with respect to \( y \), treat \( x \) as constant:

• \( f_y(x, y) = \frac{\partial}{\partial y}(2y^2 + x^2 - x^2y) = 4y - x^2 \)

By setting these partial derivatives to zero, we find critical points - values of \( x \) and \( y \) where changes in each variable do not affect the function's value. This step is crucial for determining where the function has potential maxima, minima, or saddle points.

Hessian Determinant

The Hessian determinant is part of a method to classify critical points and determine their nature, whether they are maxima, minima, or saddle points. The function's second partial derivatives come together to form a matrix called the Hessian matrix.

For our given function, we calculate:

• \( f_{xx} = 2 - 2y \)
• \( f_{yy} = 4 \)
• \( f_{xy} = -2x \)

The Hessian determinant \( H \) is then \[ H = f_{xx}f_{yy} - (f_{xy})^2 \]Using this formula, we plug in our second partial derivatives to evaluate the determinant at each critical point. The value of \( H \) tells us significant information:

• If \( H > 0 \) and \( f_{xx} > 0 \), we have a local minimum.
• If \( H > 0 \) and \( f_{xx} < 0 \), we have a local maximum.
• If \( H < 0 \), the point is a saddle point.
• If \( H = 0 \), the test is inconclusive.

Relative Maximum and Minimum

Finding relative maximum and minimum values involves analyzing where a function attains highest or lowest points in its neighborhood. For the function \( f(x, y) = 2y^2 + x^2 - x^2y \), critical points are first identified by setting the first partial derivatives to zero.

In our problem, the critical point \((0, 0)\) with a Hessian determinant value of 8 (\( H > 0\)) and \( f_{xx} > 0\), indicates a relative minimum.

• This means in the immediate vicinity, the function values are higher than \( f(0,0)\), confirming it's a minimum point.

Critical points mark locations where the slope (change) of the function is 0 in all directions, suggesting the possibility of turning points. Real-life analogies include the lowest point in a valley or peak on a hill.

Relative extremum points are powerful tools for understanding the behavior of complex functions.

Saddle Points

Saddle points are unique and interesting features of multivariable functions. Unlike maxima or minima, saddle points exhibit different curvatures in different directions.

In our function, the critical point \((2, 1)\) is classified as a saddle point owing to its negative Hessian determinant value of -16.

• This implies the function behaves like a saddle here – it goes up in one direction and down in another.
• Saddle points create a context without a local maximum or minimum but instead a different kind of inflection.

They are named for their resemblance to a typical horse saddle: there is a rise towards the front and back, but a dip along the middle.

Recognizing saddle points is vital when sketching or interpreting the graph of a function, offering insights that diverge from typical peak and valley scenarios.