Problem 22 - Calculus with Concepts in Calculus

Question

Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ k(u, v)=(u+v)^{2} $$

Step-by-Step Solution

Answer

The critical points are on the line \(u+v=0\); their nature can't be determined by the second derivative test.

Step 1: Find the First Partial Derivatives

To find the critical points, we first need to compute the partial derivatives of the function \(k(u, v) = (u+v)^2\). The partial derivative with respect to \(u\) is \(k_u = 2(u+v)\), and the partial derivative with respect to \(v\) is \(k_v = 2(u+v)\).

Step 2: Find Critical Points

To find the critical points, set the partial derivatives to zero. Solve the equations: \(2(u+v) = 0\) and \(2(u+v) = 0\). Both equations imply that \(u+v = 0\). Thus, the critical points lie on the line \(u + v = 0\).

Step 3: Compute the Second Partial Derivatives

Find the second partial derivatives: \(k_{uu} = \frac{\partial}{\partial u}(2(u+v)) = 2\), \(k_{vv} = \frac{\partial}{\partial v}(2(u+v)) = 2\), and \(k_{uv} = k_{vu} = \frac{\partial}{\partial v}(2(u+v)) = 2\).

Step 4: Determine the Nature of Each Critical Point

We use the second derivative test, which involves the Hessian matrix \(H = \begin{bmatrix} k_{uu} & k_{uv} \ k_{uv} & k_{vv} \end{bmatrix} = \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix}\). The determinant of \(H\) is \(\det(H) = (2)(2) - (2)(2) = 0\). A determinant of zero means that the test is inconclusive, thus any critical point cannot be classified as a maximum, minimum, or saddle point using this test.

Key Concepts

First Partial Derivatives

When studying functions of several variables, like our function \(k(u, v) = (u+v)^2\), partial derivatives are crucial for understanding changes in the function's output. Partial derivatives \(k_u\) and \(k_v\) represent the rates of change of \(k\) with respect to \(u\) and \(v\) respectively. To find them, we differentiate \(k\) while treating other variables as constants.

• For \(k_u\): Differentiate \((u+v)^2\) with respect to \(u\). It's like asking, "How does \(k\) change as \(u\) changes, keeping \(v\) constant?" The result is \(k_u = 2(u+v)\).
• For \(k_v\): Differentiate \( (u+v)^2 \) with respect to \(v\), treating \(u\) as constant. This gives us \(k_v = 2(u+v)\).

Both derivatives highlight how intertwined \(u\) and \(v\) are in impacting the function \(k\). Setting these partial derivatives to zero helps us identify critical points, where the function's rate of increase or decrease halts momentarily.

Second Derivative Test

The second derivative test is a handy tool for determining the nature of critical points. It involves assessing the second partial derivatives, which provide information about the curvature of the function.

• If you find yourself with a positive curvature (where second derivatives are positive), you're sitting at a relative minimum.
• Negative curvature indicates a relative maximum.
• If the test is inconclusive, like in this case, it suggests that the critical point has a more complex nature or that a simple categorization as a maximum, minimum, or saddle point isn't possible.

For our function, the determinant of the Hessian matrix turned out to be zero, meaning this conventional test doesn't help us much. When this happens, it might be necessary to explore other approaches or deeper insights to classify the critical points.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a function. It's represented as \( H = \begin{bmatrix} k_{uu} & k_{uv} \ k_{vu} & k_{vv} \end{bmatrix} \). This matrix plays a vital role in the second derivative test.For our function \(k(u, v) = (u+v)^2\), the Hessian matrix is calculated using:

• \(k_{uu} = 2\) - This indicates how the function curves with respect to \(u\).
• \(k_{vv} = 2\) - This shows the curvature with respect to \(v\).
• \(k_{uv} = k_{vu} = 2\) - These cross-partial derivatives confirm the interconnectedness of the variables \(u\) and \(v\).

The determinant of the Hessian \( \det(H) = 0 \) tells us that our Hessian matrix doesn't provide sufficient evidence to categorize the critical points using the second derivative test. It's a reminder that some critical points can behave in complex ways, sometimes beyond simple classification.