Problem 20 - Calculus Early Transcendentals

Question

Show that \(f(x, y)=x^{2} y e^{-x^{2}-y^{2}}\) has maximum values at \((\pm 1,1 / \sqrt{2})\) and minimum values at \((\pm 1,-1 / \sqrt{2}) .\) Show also that \(f\) has infinfinitely many other critical points and \(D=0\) at each of them. Which of them give rise to maximum values? Minimum values? Saddle points?

Step-by-Step Solution

Answer

The function has maximum points at \((\pm 1, \frac{1}{\sqrt{2}})\), minimum points at \((\pm 1, -\frac{1}{\sqrt{2}})\), and saddle points elsewhere.

Step 1: Find Partial Derivatives

First, find the first partial derivatives of the function with respect to both \(x\) and \(y\). The function is \(f(x, y) = x^2 y e^{-x^2-y^2}\).- The partial derivative with respect to \(x\) is: \[f_x = \frac{\partial}{\partial x}(x^2 y e^{-x^2-y^2}) = 2xy e^{-x^2-y^2} - 2x^3 y e^{-x^2-y^2}\] Simplify this: \[f_x = 2xy e^{-x^2-y^2}(1 - x^2)\]- The partial derivative with respect to \(y\) is: \[f_y = \frac{\partial}{\partial y}(x^2 y e^{-x^2-y^2}) = x^2 e^{-x^2-y^2} - 2x^2 y^2 e^{-x^2-y^2}\] Simplify this: \[f_y = x^2 e^{-x^2-y^2}(1 - 2y^2)\]

Step 2: Find Critical Points

To find the critical points, set the partial derivatives equal to zero:1. From \(f_x = 0\): \[2xy(1 - x^2) = 0\] This gives conditions: (i) \(x = 0\) or (ii) \(y = 0\) or (iii) \(x^2 = 1\).2. From \(f_y = 0\): \[x^2(1 - 2y^2) = 0\] This gives conditions: (i) \(x = 0\) or (ii) \(y^2 = \frac{1}{2}\).Combining these: The critical points are \((0, y)\), \((x, 0)\), and \((\pm 1, \pm \frac{1}{\sqrt{2}})\).

Step 3: Evaluate the Second Derivatives

To determine the nature of the critical points, we need the second derivatives:- \(f_{xx} = \frac{\partial}{\partial x}(2xy(1 - x^2)) = 2y(1 - 3x^2)\)- \(f_{yy} = \frac{\partial}{\partial y}(x^2(1 - 2y^2)) = -4x^2 y\)- \(f_{xy} = \partial_x \partial_y (f) = 2x(1 - x^2) - 4xy^2\)\- Since \(f_{xy} = 2x(1-x^2)\) (after simplifying again), notice the symmetry in derivatives.

Step 4: Determine the Discriminant

The discriminant \(D\) is given by:\[D = f_{xx} f_{yy} - (f_{xy})^2\]Evaluate it at each critical point:1. For points \((0, y)\), \(D=0\), since \(f_{xx}\), \(f_{yy}\), \(f_{xy}=0\).2. For points \((\pm 1, \pm \frac{1}{\sqrt{2}})\), calculate using values of partial second derivatives at those points (e.g., \(f_{xx}(1, 1/\sqrt{2})\)).

Step 5: Classify Critical Points

For maximum/minimum classification:1. \((\pm 1, \frac{1}{\sqrt{2}})\): - check \(D > 0\) and \(f_{xx} < 0\), implies maximum.2. \((\pm 1, -\frac{1}{\sqrt{2}})\): - similarly, checks \(D > 0\), \(f_{xx} > 0\), implies minimum.3. Other critical points: - \((0, y)\) are evaluated with respect to \(D=0\), classify based on further second derivative tests (often saddle points).

Key Concepts

Partial Derivatives

Partial derivatives are vital in multivariable calculus, as they allow us to study how a function changes with respect to one variable at a time, while keeping other variables constant. When dealing with the function \( f(x, y) = x^2 y e^{-x^2 - y^2} \), we calculate partial derivatives to identify how it changes with respect to \( x \) and \( y \).

The partial derivative with respect to \( x \), denoted as \( f_x \), is found by differentiating the function while treating \( y \) as a constant:

• \( f_x = 2xy e^{-x^2 - y^2} (1 - x^2) \)

Similarly, the partial derivative with respect to \( y \), \( f_y \), is:

• \( f_y = x^2 e^{-x^2 - y^2} (1 - 2y^2) \)

These derivatives help in locating critical points by setting them equal to zero.

Critical Points

Critical points provide crucial insights into the behavior of a function. They occur where the gradient is zero, meaning both partial derivatives are simultaneously zero. For our function \( f(x, y) \), the conditions for critical points come from:

• \( f_x = 0 \Rightarrow 2xy(1 - x^2) = 0 \)
• \( f_y = 0 \Rightarrow x^2(1 - 2y^2) = 0 \)

Solving these equations gives us possible values of \( x \) and \( y \) as critical points:

• \( (0, y), (x, 0), (\pm 1, \pm \frac{1}{\sqrt{2}}) \)

These points need further analysis, using the second derivative test, to determine their nature.

Second Derivative Test

The second derivative test helps classify critical points as maxima, minima, or saddle points. We compute second-order partial derivatives:

• \( f_{xx} = 2y (1 - 3x^2) \)
• \( f_{yy} = -4x^2 y \)
• \( f_{xy} = 2x(1 - x^2) \)

These derivatives form a matrix known as the Hessian matrix of second derivatives. At each critical point, we'll determine the discriminant, \( D \), to classify the point. This test helps distinguish between maximum, minimum, and saddle points based on the values of \( f_{xx} \), \( f_{yy} \), and \( f_{xy} \).

Discriminant

The discriminant in multivariable calculus is a determinant used widely to classify critical points. For the given function \( f(x, y) \), it is expressed as: \[ D = f_{xx} f_{yy} - (f_{xy})^2 \]It’s used to determine the nature of critical points:

• \( D > 0 \) and \( f_{xx} < 0 \): local maximum
• \( D > 0 \) and \( f_{xx} > 0 \): local minimum
• \( D < 0 \): saddle point
• At some points, if \( D = 0 \), this can indicate more complex behaviors like higher order tests needed

For the function \( f(x, y) \), evaluating \( D \) helps us decipher whether the critical points are maxima, minima, or saddle points.

Saddle Points

Saddle points are fascinating points where the function neither has a maximum nor a minimum. They occur when the discriminant \( D < 0 \). This means the function curves upwards in one direction and downwards in another, resembling a saddle.

When checking our critical points for \( f(x, y) \), especially those with \( D = 0 \) at points like \((0, y)\), we need further exploration beyond the basic second derivative test. Often, these points exhibit saddle behavior due to the nature of the function's curvature at these locations. By analyzing these points, we grasp the overall geometry of the function's graph.