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Chapter 14: Problem 8

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$ f(x, y)=y\left(e^{x}-1\right) $$

Short Answer

Expert verified
Saddle point at (0, 0); no local maxima or minima.

Step by step solution

01

Find the Partial Derivatives

To identify critical points, we first need the partial derivatives of the function. The function is \( f(x, y) = y(e^x - 1) \). Calculate the partial derivatives: 1. With respect to \( x \):\[ f_x(x, y) = \frac{\partial}{\partial x} [y(e^x - 1)] = y e^x \]2. With respect to \( y \):\[ f_y(x, y) = \frac{\partial}{\partial y} [y(e^x - 1)] = e^x - 1 \].
02

Set Partial Derivatives to Zero

To find critical points, set both partial derivatives equal to zero and solve for \( x \) and \( y \):1. \( f_x = y e^x = 0 \) implies either \( y = 0 \) or \( e^x = 0 \). Since \( e^x \) cannot be zero, we have \( y = 0 \).2. \( f_y = e^x - 1 = 0 \) implies \( e^x = 1 \), which means \( x = 0 \) because \( e^0 = 1 \).
03

Determine Critical Points

The solutions where both equations from Step 2 are satisfied simultaneously gives us the critical points. When \( y = 0 \) and \( x = 0 \), we have the critical point \((x, y) = (0, 0)\).
04

Second Partial Derivatives Test

To determine the nature of the critical points, compute the second partial derivatives and use the Hessian matrix:1. Second partial derivatives: - \( f_{xx} = \frac{\partial}{\partial x}[y e^x] = y e^x \) - \( f_{yy} = \frac{\partial}{\partial y}[e^x - 1] = 0 \) - \( f_{xy} = \frac{\partial}{\partial y}[y e^x] = e^x \)2. Evaluate the Hessian at the critical point \((0, 0)\): - \( f_{xx}(0, 0) = 0 \), \( f_{yy}(0, 0) = 0 \), \( f_{xy}(0, 0) = 1 \).3. Hessian determinant: \( H = f_{xx} f_{yy} - (f_{xy})^2 = 0 \times 0 - (1)^2 = -1 \).The negative determinant indicates a saddle point at \((0, 0)\).
05

Conclusion

The function \( f(x, y) = y(e^x - 1) \) has one saddle point at \((0, 0)\). There are no local maxima or minima.

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Key Concepts

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

Partial Derivatives
In mathematical analysis, finding partial derivatives is a crucial step when dealing with functions of multiple variables. For a function like \( f(x, y) = y(e^x - 1) \), you need to understand how it changes with respect to each variable separately 鈥� in this case, \( x \) and \( y \).
To find the partial derivative with respect to \( x \), treat \( y \) as a constant and differentiate the expression \( y(e^x - 1) \). This gives \( f_x = y e^x \). Similarly, for \( f_y \), treat \( x \) as constant, leading to \( f_y = e^x - 1 \).
The partial derivatives provide insight into the slope of the tangent plane at any point on the surface described by \( f \), indicating how the function is changing at that point in space.
Critical Points
Critical points are locations on the graph of a function where the gradient (or first derivative) is zero or undefined. To locate these for a function like \( f(x, y) \), set both partial derivatives from the previous section to zero:
  • \( f_x = y e^x = 0 \)
  • \( f_y = e^x - 1 = 0 \)
For \( f_x = 0 \), either \( y = 0 \) or \( e^x = 0 \). Since \( e^x \) is never zero, we conclude \( y = 0 \). From \( f_y = 0 \), since \( e^x = 1 \) equals \( e^0 \), we find \( x = 0 \). Thus, the solution \((x, y) = (0, 0)\) is our critical point.
These are essential because they help us identify the nature of the function around these points - whether it is at a peak (local maximum), valley (local minimum), or a saddle (neither).
Second Partial Derivatives Test
After locating critical points, applying the second partial derivatives test helps determine their nature. This involves computing the second order partial derivatives and organizing them into what is known as the Hessian matrix. For our function, these are calculated as follows:
  • \( f_{xx} = y e^x \)
  • \( f_{yy} = 0 \)
  • \( f_{xy} = e^x \)
The Hessian matrix formulated from these derivatives is evaluated at the critical point \((0, 0)\), yielding:
  • \( f_{xx}(0, 0) = 0 \)
  • \( f_{yy}(0, 0) = 0 \)
  • \( f_{xy}(0, 0) = 1 \)
The determinant of this matrix, \( H = f_{xx} f_{yy} - (f_{xy})^2 = -1 \), indicates a negative value, signifying a saddle point at the origin \((0, 0)\).
Thus, the Second Partial Derivatives Test effectively tells us the nature of our critical point, confirming that there are no maximum or minimum values, but rather a saddle point.

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Most popular questions from this chapter

Show that the operation of taking the gradient of a function has the given property. Assume that \(u\) and \(v\) are differentiable functions of \(x\) and \(y\) and that \(a, b\) are constants. (a) \(\nabla(a u+b v)=a \nabla u+b \nabla v\) (b) \(\nabla(u v)=u \nabla v+v \nabla u\) (c) \(\nabla\left(\frac{u}{v}\right)=\frac{v \nabla u-u \nabla v}{v^{2}}\) (d) \(\nabla u^{n}=n u^{n-1} \nabla u\)

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