Problem 38 - Calculus Early Transcendentals

Question

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.

Step-by-Step Solution

Answer

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

Step 1: Find the First-Order Partial Derivatives

To find the critical points of the function, compute the first-order partial derivatives of the function \(f(x, y) = 3xe^y - x^3 - e^{3y}\). This gives:\[ f_x = \frac{\partial}{\partial x}(3xe^y - x^3 - e^{3y}) = 3e^y - 3x^2 \]\[ f_y = \frac{\partial}{\partial y}(3xe^y - x^3 - e^{3y}) = 3xe^y - 3e^{3y} \]

Step 2: Set Partial Derivatives to Zero and Solve

To find critical points, solve the equations obtained by setting the partial derivatives to zero:\[ 3e^y - 3x^2 = 0 \]\[ 3xe^y - 3e^{3y} = 0 \]Solving the first equation, we get:\[ e^y = x^2 \]Using this in the second equation yields:\[ 3x(x^2) - 3(x^2)^3 = 0 \]\[ 3x^3 - 3x^6 = 0 \]\[ 3x^3(1 - x^3) = 0 \]This gives the solutions \(x = 0\) and \(x = 1\). Test only \(x = 1\). Thus, for \(x = 1\), \(e^y = 1^2 = 1\), hence \(y = 0\). So the critical point is \((x, y) = (1, 0)\).

Step 3: Classify the Critical Point Using the Hessian Matrix

Next, we calculate the second-order partial derivatives:\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = -6x \]\[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 3xe^y - 9e^{3y} \]\[ f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = 3e^y \]At \((x, y) = (1, 0)\), evaluate the derivatives:\[ f_{xx} = -6 \quad, \quad f_{yy} = 3 \times 1 \times e^0 - 9 \times e^0 = -6 \quad, \quad f_{xy} = 3 \times e^0 = 3\]Calculate the determinant of the Hessian:\[ D = f_{xx}f_{yy} - (f_{xy})^{2} = (-6)(-6) - (3)^{2} = 36 - 9 = 27 \]Since \(D > 0\) and \(f_{xx} < 0\), the point \((1,0)\) is a local maximum.

Step 4: Evaluate Function at Critical Point and Boundary

Evaluate \(f(1,0)\):\[ f(1,0) = 3 \times 1 \times e^0 - 1^3 - e^{0} = 3 - 1 - 1 = 1 \]

Step 5: Find Points with Higher Values: Establishing Not an Absolute Maximum

Evaluate the function at different points, such as along (x, y) = (t, log(t^2)) outside the critical area:\[ f(t, \ln(t^2)) = 3t(t^2) - t^3 - e^{3*ln(t^2)} = 2t^3 - t^6 \]Choose t = 2, which yields:\[ f(2, \ln(4)) = 2(2)^3 - 2^6 = 16 - 64 = -48 \]This suggests that as \(x\) increases further, the function value is greater than at (1,0). Hence, (1,0) is not an absolute maximum.

Step 6: Visualize the Function

To fully illustrate the situation, employ a computer graphing tool to plot the function over a domain such as \([-2, 2]\) for \(x\) and \([-1, 1]\) for \(y\). Choose a perspective that clearly shows the peak at \((1, 0)\) as a local maximum but shows increasing values away from it.

Key Concepts

Critical Points

In multivariable calculus, a critical point is a location on a function where the gradient (or the vector of all first-order partial derivatives) is zero. For functions of two variables, such as our function \(f(x, y) = 3xe^y - x^3 - e^{3y}\), critical points can indicate the function's peaks, valleys, or saddle points.
To find these points, set the partial derivatives with respect to both variables to zero. Here, we compute:

• \( f_x = 3e^y - 3x^2 \)
• \( f_y = 3xe^y - 3e^{3y} \)

By solving these equations, we found one critical point at \((1, 0)\).
This means at this coordinate, the function changes behavior, and further analysis is required to classify this point.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a function. It plays a pivotal role in optimizing multivariable functions and classifying critical points.
For our function \(f(x, y)\), the Hessian matrix is:

• \( f_{xx} = -6x \)
• \( f_{yy} = 3xe^y - 9e^{3y} \)
• \( f_{xy} = f_{yx} = 3e^y \)

At the critical point \((1, 0)\), the Hessian determinant is \(D = f_{xx}f_{yy} - (f_{xy})^2 = 27\).
Since \(D > 0\) and \(f_{xx} < 0\), the critical point is classified as a local maximum.

Local Maximum

A local maximum occurs at a point if the function's value is greater than at any nearby point. To determine if a critical point like \((1, 0)\) is a local maximum, use the Hessian test.
At \((1, 0)\), we classified it as a local maximum because:
- The Hessian determinant \(D = 27 > 0\)
- The second derivative \(f_{xx} = -6 < 0\)
This confirms the spot as a local peak on the function's graph.
But remember, local doesn't mean global. For instance, along certain paths like \((t, \ln(t^2))\), the function shows higher values. This underscores that \((1,0)\) is not the absolute peak over the entire domain.

Partial Derivatives

Partial derivatives measure how a function changes as only one of the variables is altered, keeping the others constant.
For the function \(f(x, y)\), the partial derivatives are:

• \( f_x = \frac{\partial f}{\partial x} = 3e^y - 3x^2 \)
• \( f_y = \frac{\partial f}{\partial y} = 3xe^y - 3e^{3y} \)

These derivatives at specific points show the function's rate of change along each axis.
Setting them to zero helps pinpoint the critical points. Here, solving the equations \(3e^y = 3x^2\) and \(3xe^y = 3e^{3y}\) led to finding \((x, y) = (1, 0)\) as a critical point.