Problem 36 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. Suppose \(h(x, y)=f(x)+g(y)\), where \(f\) and \(g\) have continuous second derivatives near \(a\) and \(b\), respectively. If \(a\) is a critical number of \(f, b\) is a critical number of \(g\), and \(f^{\prime \prime}(a) g^{\prime \prime}(b)>0\), then \(h\) has a relative extremum at \((a, b)\).

Step-by-Step Solution

Answer

The statement is false. Even if \(f''(a)g''(b) > 0\), it does not automatically mean \(h(x, y)\) will have a relative extremum at \((a, b)\). For example, if \(f(x) = x^2, g(y) = -y^2, h(x, y) = x^2 - y^2\), then \(a = b = 0\) are critical points of \(f\) and \(g\), and \(f''(a)g''(b) = (2)(-2) = -4 > 0\). However, \(h(x, y)\) does not have a relative extremum at \((0, 0)\); instead, it has a saddle point there, which indicates that these conditions are not sufficient to ensure a relative extremum for \(h\).

Step 1: Restating the function h(x, y)

We know that \(h(x, y) = f(x) + g(y)\).

Step 2: Analysis of critical numbers and second derivatives of f and g

We've got that \(f''(a)\) and \(g''(b)\) exist, which means they are differentiable. As \(a\) is a critical number of \(f\) and \(b\) is a critical number of \(g\), we know that \(f'(a) = 0\) and \(g'(b) = 0\).

Step 3: Second Derivative Test for relative extremum

The Second Derivative Test for a one-variable function says that if \(f'(a) = 0\) and \(f''(a) > 0\), then \(f\) has a local minimum at \(a\). Similarly, if \(g'(b) = 0\) and \(g''(b) > 0\), then \(g\) has a local minimum at \(b\). The supposition \(f''(a)g''(b) > 0\) should mean that \(f''(a)\) and \(g''(b)\) are both positive or both negative, which indicates local minimum points or local maximum points for \(f\) and \(g\), respectively.

Step 4: Verify the relative extremum of h at (a, b)

Even though \(f\) has a local extremum at \(a\) and \(g\) has a local extremum at \(b\), it doesn't necessarily mean \(h(x, y)\) will have a relative extremum at \((a, b)\), because \(h\) is the sum of those two functions and their critical points are not related. The results from each may cancel each other out or not, depending on their exact values.

Step 5: Conclusion

The statement is therefore false: Even if \(f''(a)g''(b) > 0\), it does not automatically mean \(h(x, y)\) will have a relative extremum at \((a, b)\). For example, if \(f(x) = x^2, g(y) = -y^2, h(x, y) = x^2 - y^2\), then \(a = b = 0\) are critical points of \(f\) and \(g\), and \(f''(a)g''(b) = (2)(-2) = -4 > 0\). However, \(h(x, y)\) does not have a relative extremum at \((0, 0)\); instead, it has a saddle point there, which indicates that these conditions are not sufficient to ensure a relative extremum for \(h\).

Key Concepts

Critical Points

In multivariable calculus, when we look for critical points of a function of several variables, we examine where the partial derivatives are equal to zero.
These are potential spots where the function can have a local minimum, maximum, or a saddle point.
Specifically, for a function like \(h(x, y) = f(x) + g(y)\), we need to find the critical points for \(f\) and \(g\) individually. Here's how it works:

• Find where the first derivative \(f'(x)\) is zero. This gives a critical point \(a\) for \(f(x)\).
• Similarly, find where \(g'(y)\) is zero to obtain a critical point \(b\) for \(g(y)\).

Generally, if both derivatives are zero, the point \((a, b)\) could be a critical point for \(h(x, y)\).
However, this doesn't ensure it’s a relative extremum without further analysis.

Second Derivative Test

The Second Derivative Test helps determine the nature of a critical point.
It analyzes the second derivatives to decide whether we have a minimum, maximum, or saddle point.For one-variable functions like \(f(x)\) and \(g(y)\):

• If \(f''(a) > 0\), \(f(x)\) has a local minimum at \(x = a\).
• If \(f''(a) < 0\), \(f(x)\) has a local maximum at \(x = a\).
• Similarly, \(g''(b) > 0\) or \(g''(b) < 0\) tell us about \(g(y)\) at \(y = b\).

But combining these isn't straightforward for \(h(x, y) = f(x) + g(y)\).
The inequality \(f''(a)g''(b) > 0\) means both second derivatives are either positive or both negative,
suggesting both contribute to a kind of extremum.
Yet, due to the nature of \(h\) being a sum, it might not lead necessarily to a clear extremum, as their influences on \(h\) may cancel out.

Relative Extremum

A relative extremum is where a function takes a mini-peak (relative maximum) or mini-trough (relative minimum).
It is relative because it’s only compared to its immediate surroundings, not the entire domain.For \(h(x, y) = f(x) + g(y)\), examining critical points \((a, b)\) and using the Second Derivative Test:

• Suppose \(f''(a) > 0\) and \(g''(b) > 0\), \(h\) might seem to have a relative minimum at \((a, b)\).
• Conversely, if both are less than zero, you might expect a relative maximum.

However, because \(h\) is the sum, every small behavior might balance out, leading to neither a max nor a min.
In fact, as shown in the example from the given solution,
even when \(f''(a)g''(b) > 0\), \(h(x, y)\) had a saddle point at \((0,0)\), illustrating these conditions weren't sufficient for an extremum.
Therefore, these tests in multivariable cases should be supplemented by examining other aspects or understanding of the function's behavior.