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Chapter 11: Problem 30

If a function of one variable is continuous on an interval and has only one critical number, then a local maximum has to be an absolute maximum. But this is not true for functions of two variables. Show that the function $$f(x, y)=3 x e^{y}-x^{3}-e^{3 y}$$ has exactly one critical point, and that \(f\) has a local maximum there that is not an absolute maximum. Then use a computer to produce a graph with a carefully chosen domain and viewpoint to see how this is possible.

Short Answer

Expert verified
The function has one critical point at \( (1, 0) \), which is a local maximum but not an absolute maximum.

Step by step solution

01

Determine the Partial Derivatives

To find the critical points of the function \( f(x, y) = 3x e^y - x^3 - e^{3y} \), we first need to find the partial derivatives with respect to \( x \) and \( y \). The partial derivative with respect to \( x \) is:\[ f_x = \frac{\partial}{\partial x}(3x e^y - x^3 - e^{3y}) = 3e^y - 3x^2. \]The partial derivative with respect to \( y \) is:\[ f_y = \frac{\partial}{\partial y}(3x e^y - x^3 - e^{3y}) = 3xe^y - 3e^{3y}. \]
02

Solve for Critical Points

To find critical points, we set the partial derivatives equal to zero:1. \( f_x = 3e^y - 3x^2 = 0 \)2. \( f_y = 3xe^y - 3e^{3y} = 0 \)Solving these equations, we begin with \( 3e^y = 3x^2 \) which simplifies to \( e^y = x^2 \).Substitute \( e^y = x^2 \) into the second equation:\[ 3x(x^2) - 3(x^2)^3 = 0 \]\[ 3x^3 - 3x^6 = 0 \]\[ 3x^3(1 - x^3) = 0. \]This yields solutions \( x = 0 \) or \( 1 - x^3 = 0 \). From \( 1 - x^3 = 0 \), we get \( x = 1 \). Therefore, \( x = 0 \) or \( x = 1 \).
03

Determine the Point for x = 1

For \( x = 1 \), solving \( e^y = x^2 = 1 \) gives \( y = 0 \) since \( e^0 = 1 \). Thus, one critical point is \( (1, 0) \).
04

Verify No Critical Point at x = 0

For \( x = 0 \), since \( e^y = x^2 = 0 \), it results in a contradiction because \( e^y \) cannot equal zero, thus no critical point exists when \( x = 0 \). So the only critical point is \( (1, 0) \).
05

Check for Local Maximum or Minimum

We check the nature of the critical point by considering the second-order partial derivatives of \( f \):- \( f_{xx}= -6x \)- \( f_{yy}= -9e^{3y} \)- \( f_{xy}= 3e^y \)Using the Hessian matrix, \( H = f_{xx}f_{yy} - (f_{xy})^2 \).At \( (1, 0) \):\[ H = (-6)(-9) - (3)^2 = 54 - 9 = 45. \]Since \( H > 0 \) and \( f_{xx} < 0 \), \( (1, 0) \) is a local maximum.
06

Show it is Not an Absolute Maximum

Evaluate \( f(x, y) \) as \( y \to -\infty \), then \( e^y \to 0 \) and \( e^{3y} \to 0 \), so the function dominates by the cubic term \( -x^3 \). Hence, as \( x \to \infty \) or \( x \to -\infty \), \( f(x, y) \to -\infty \), demonstrating that the local maximum \( (1, 0) \) is not the topmost point of the function. Thus, not an absolute maximum.

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Key Concepts

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

Critical Points
Critical points are where the derivative of a function is zero or undefined, and they provide valuable information about the function's behavior. For functions of two variables, critical points occur at the values of
  • \( x \) and \( y \) where all partial derivatives become zero or are undefined.
  • These points can signal local maxima, minima, or saddle points, depending on the nature of the function around them.
To find critical points for the function \( f(x, y) = 3x e^y - x^3 - e^{3y} \), we compute the partial derivatives with respect to \( x \) and \( y \), then set them to zero. This process reveals the critical point at \( (1, 0) \). Understanding critical points helps one analyze where changes in a function are most intense and direct, crucial for studying complex systems.
Partial Derivatives
Partial derivatives are fundamental when working with functions of multiple variables. They measure the rate at which a function changes as
  • One variable changes while all other variables are held constant.
  • For a function \( f(x, y) \), we find the partial derivative with respect to \( x \) by treating \( y \) as fixed and differentiating with respect to \( x \), and vice versa for \( y \).
In our example, the partial derivatives \( f_x = 3e^y - 3x^2 \) and \( f_y = 3xe^y - 3e^{3y} \) are derived. They allow us to
  • Identify the slope and rate of change for the function along the x- and y-axes, respectively.
  • They provide a pathway to calculate and solve for critical points.
By mastering partial derivatives, students gain essential tools for analyzing surfaces and changes in multi-variable systems.
Local Maximum
A local maximum is a point on a function where the value is higher than all nearby points, indicating a peak in the function. Identifying a local maximum
  • Involves checking not just the function value at the critical point, but also the concavity of the function using the second-order partial derivatives.
  • For the function at \( (1, 0) \), evaluating the Hessian matrix helps confirm its nature.
  • If the Hessian is positive at a critical point, and \( f_{xx} \) is negative, it indicates a local maximum.
  • This was shown to be the case for this function at \( (1, 0) \), but it isn't an absolute maximum because the function's values drop off to \(-\infty\) as \( x \) and \( y \) trend away from this point.
Understanding local maxima is vital, as it helps determine potential peak values within regions of a domain, offering insights into optimization and function behaviors.

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Most popular questions from this chapter

The plane \(4 x-3 y+8 z=5\) intersects the cone \(z^{2}=x^{2}+y^{2}\) in an ellipse. (a) Graph the cone and the plane, and observe the resulting ellipse. (b) Use Lagrange multipliers to find the highest and lowest points on the ellipse.

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. $$f(x, y)=e^{x} \cos y$$

Use Lagrange multipliers to prove that the triangle with maximum area that has a given perimeter \(p\) is equilateral. Hint: Use Heron's formula for the area: $$ A=\sqrt{s(s-x)(s-y)(s-z)} $$ where \(s=p / 2\) and \(x, y, z\) are the lengths of the sides.

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. $$ f(x, y)=e^{x y} ; \quad x^{3}+y^{3}=16 $$

Find the dimensions of a rectangular box of maximum volume such that the sum of the lengths of its 12 edges is a constant \(c .\)
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