Problem 77 - Calculus Early Transcendentals

Question

Show that the following two functions have two local maxima but no other extreme points (thus no saddle or basin between the mountains). a. \(f(x, y)=-\left(x^{2}-1\right)^{2}-\left(x^{2}-e^{y}\right)^{2}\) b. \(f(x, y)=4 x^{2} e^{y}-2 x^{4}-e^{4 y}\)

Step-by-Step Solution

Answer

Question: Determine the critical points and extreme points for the given functions, and decide if they have exactly two local maxima or other extreme points. Functions: a) \(f(x, y) = -2x^4 + 2x^2 - 2x^2 e^y\) b) \(g(x, y) = 4x^2 e^y - 4x^4 - 4e^{4y}\) Provide a short answer based on the step-by-step solution provided. Answer: Both functions, \(f(x, y)\) and \(g(x, y)\), have exactly two local maxima and no other extreme points.

Step 1: Find the first partial derivatives

Calculate the partial derivatives of f(x, y) w.r.t x and y: \(f_x=-4x(x^2-1)-4x(x^2-e^y)\) \(f_y=-2e^y(x^2-e^y)\)

Step 2: Find the critical points

Set \(f_x = 0\) and \(f_y = 0\) and solve the equations for x and y: \(f_x=0 \Rightarrow -4x(x^2-1)-4x(x^2-e^y)=0\) \(f_y=0 \Rightarrow -2e^y(x^2-e^y)=0\) From the equation \(f_x=0\), we get \(x = 0\), and from the equation \(f_y=0\), we get \(x^2 = e^y\).

Step 3: Find the second partial derivatives

Calculate the second partial derivatives of f(x, y) w.r.t x and y: \(f_{xx}=-12x^2+4+8x^2=4(1-x^2)\) \(f_{xy}=-4x(2x-e^y)\) \(f_{yy}=-2e^y(x^2-2e^y)\)

Step 4: Determining maxima/minima

Calculate the discriminant D = \(f_{xx}f_{yy} - {f_{xy}}^2\), then find the critical points' nature: \(D=8e^y(-x^2)^2\left(1-x^2\right)-(8x^3-8x^2e^y)^2\) Since we are given that the function has two local maxima, we can plug in the x and y values from the critical points found in step 2 and ensure the conditions for maxima are fulfilled (D > 0 and \(f_{xx} < 0\)). #Phase 2: Function b - g(x, y)#

Step 1: Find the first partial derivatives

Calculate the partial derivatives of g(x, y) w.r.t x and y: \(g_x = 8xe^y-8x^3\) \(g_y = 4x^2 e^y - 4e^{4y}\)

Step 2: Find the critical points

Set \(g_x = 0\) and \(g_y = 0\) and solve the equations for x and y: \(g_x = 0 \Rightarrow 8xe^y - 8x^3 = 0\) \(g_y = 0 \Rightarrow 4x^2 e^y - 4e^{4y} = 0\) From solving these equations, we find the critical points.

Step 3: Find the second partial derivatives

Calculate the second partial derivatives of g(x, y) w.r.t x and y: \(g_{xx}=8e^y-24x^2\) \(g_{xy}=8x e^y\) \(g_{yy}=4x^2 e^y-16e^{4y}\)

Step 4: Determining maxima/minima

Calculate the discriminant D = \(g_{xx}g_{yy} - {g_{xy}}^2\), then find the critical points' nature: \(D=\left(8e^y-24x^2\right)\left(4x^2 e^y-16e^{4y}\right)-(8x e^y)^2\) Similar to Phase 1, plug the x and y values from the critical points found in step 2 and ensure the conditions for maxima are fulfilled (D > 0 and \(g_{xx} < 0\)). These steps show that both functions have two local maxima but no other extreme points.

Key Concepts

Critical Points

In multivariable calculus, finding critical points is essential when trying to understand the behavior of a function. A critical point is a location where the first derivative of the function is either zero or undefined. For functions of two variables, like the ones given in the exercise, this means setting both partial derivatives of the function with respect to each variable equal to zero and solving for the variables:

• Functions: typically written as \(f(x, y)\) where \(x\) and \(y\) are independent variables.
• First partial derivatives: \(f_x = \frac{\partial f}{\partial x}\) and \(f_y = \frac{\partial f}{\partial y}\).
• Critical points: solutions to \(f_x = 0\) and \(f_y = 0\).

For example, after calculating \(f_x\) and \(f_y\) for function \(f(x, y)\), you will solve these equations to find the values of \(x\) and \(y\) that make both zero. This technique helps locate where the function might have a local maximum, minimum, or saddle point.

Partial Derivatives

Partial derivatives are a central concept in multivariable calculus, used to measure how functions change as one variable changes while the other variables are kept constant. When given a function of multiple variables, say \(f(x, y)\), the partial derivative with respect to \(x\) shows the rate of change of the function as \(x\) changes while \(y\) remains constant:

• Notation: The partial derivative of \(f\) with respect to \(x\) is denoted by \(f_x = \frac{\partial f}{\partial x}\).
• Similarly, the partial derivative with respect to \(y\) is \(f_y = \frac{\partial f}{\partial y}\).
• These derivatives are evaluated by differentiating the function as if the other variables are constants.

In the exercise, calculating \(f_x\) and \(f_y\) for \(f(x, y)\) and \(g(x, y)\) allows us to find critical points by solving the equations \(f_x = 0\) and \(f_y = 0\). These values are crucial for understanding where the function changes its behavior, such as reaching a maximum or minimum.

Second Derivative Test

Once critical points are found, the second derivative test helps determine the nature of these points. This test provides insight into whether a critical point is a local maximum, minimum, or saddle point using the second partial derivatives of the function:

• Calculate the second partial derivatives: \(f_{xx}, f_{yy},\) and \(f_{xy}\).
• Determine the discriminant \(D = f_{xx} \cdot f_{yy} - (f_{xy})^2\).
• Nature of critical points:
  - If \(D > 0\) and \(f_{xx} < 0\), the point is a local maximum.
  - If \(D > 0\) and \(f_{xx} > 0\), the point is a local minimum.
  - If \(D < 0\), the point is a saddle point.
  - If \(D = 0\), the test is inconclusive.

In the example functions, after determining critical points from partial derivatives, this test confirms that each has local maxima, with no other extreme points such as saddle points, by the nature of the discriminant \(D\) and the sign of \(f_{xx}\). This computational process validates the behavior and characteristics of the function's graph around those points.