Problem 10 - 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. $$ f(u, v)=u^{3}+v^{3}-6 u v $$

Step-by-Step Solution

Answer

Critical points: (0, 0) is a saddle point, (2, 2) is a relative minimum, (-2, 2) is a saddle point.

Step 1: Find Partial Derivatives

To find the critical points, first compute the partial derivatives of \( f(u, v) = u^3 + v^3 - 6uv \). The partial derivative with respect to \( u \) is:\[ f_u = \frac{\partial f}{\partial u} = 3u^2 - 6v \]The partial derivative with respect to \( v \) is:\[ f_v = \frac{\partial f}{\partial v} = 3v^2 - 6u \]

Step 2: Set Partial Derivatives to Zero

Set the partial derivatives equal to zero to find the critical points:1. \( 3u^2 - 6v = 0 \)2. \( 3v^2 - 6u = 0 \)

Step 3: Solve the System of Equations

Solve the system of equations from Step 2:1. From the first equation: \( u^2 = 2v \) or \( v = \frac{u^2}{2} \)2. Substitute \( v = \frac{u^2}{2} \) into the second equation: \[ 3\left(\frac{u^2}{2}\right)^2 - 6u = 0 \] Simplifying gives \( \frac{3}{4}u^4 - 6u = 0 \)3. Factor the equation: \( u(u^3 - 8) = 0 \) - So, \( u = 0 \) or \( u^3 = 8 \) giving solutions: \( u = 2 \) or \( u = -2 \)

Step 4: Find Corresponding 'v' Values

Using \( u = 0 \), \( v = \frac{0^2}{2} = 0 \).Using \( u = 2 \), \( v = \frac{2^2}{2} = 2 \).Using \( u = -2 \), \( v = \frac{(-2)^2}{2} = 2 \).Critical points are: \((0, 0)\), \((2, 2)\), and \((-2, 2)\).

Step 5: Second Derivatives for the Hessian Matrix

Compute the second derivatives:- \( f_{uu} = \frac{\partial^2 f}{\partial u^2} = 6u \)- \( f_{vv} = \frac{\partial^2 f}{\partial v^2} = 6v \)- \( f_{uv} = \frac{\partial^2 f}{\partial u \partial v} = -6 \)

Step 6: Evaluate the Hessian Determinant

At each critical point, evaluate the Hessian determinant, \( H = f_{uu}f_{vv} - (f_{uv})^2 \).1. For \((0, 0)\): \( f_{uu} = 0, f_{vv} = 0 \) \( H = 0 \cdot 0 - (-6)^2 = -36 \) (saddle point)2. For \((2, 2)\): \( f_{uu} = 12, f_{vv} = 12 \) \( H = 12 \cdot 12 - (-6)^2 = 108 \) (> 0 and \(f_{uu} > 0\) implies a relative minimum)3. For \((-2, 2)\): \( f_{uu} = -12, f_{vv} = 12 \) \( H = -12 \cdot 12 - (-6)^2 = -180 \) (< 0 indicates a saddle point)

Key Concepts

Partial Derivatives

In multivariable calculus, partial derivatives are like regular derivatives but for functions with more than one variable. Instead of considering the effect of all variables, a partial derivative measures how a function changes as only one variable changes while keeping others constant.
The given function is \( f(u, v) = u^3 + v^3 - 6uv \). To find the partial derivatives, we differentiate the function first with respect to \( u \) and then with respect to \( v \). These partial derivatives help us identify the points where the function flattens out, known as critical points.
For example:

• The partial derivative with respect to \( u \) is: \( f_u = 3u^2 - 6v \).
• The partial derivative with respect to \( v \) is: \( f_v = 3v^2 - 6u \).

Finding where these derivatives equal zero is key to locating critical points.

Hessian Matrix

The Hessian Matrix is a square matrix of second-order partial derivatives of a scalar-valued function. It is crucial in classifying critical points, especially in functions with two variables. The Hessian gives us valuable information about the curvature of the function at given points.
For our function \( f(u, v) = u^3 + v^3 - 6uv \), we calculate the second partial derivatives:

• \( f_{uu} = \frac{\partial^2 f}{\partial u^2} = 6u \)
• \( f_{vv} = \frac{\partial^2 f}{\partial v^2} = 6v \)
• \( f_{uv} = \frac{\partial^2 f}{\partial u \partial v} = -6 \)

These derivatives form the Hessian Matrix. Checking the Hessian determinant \( H = f_{uu}f_{vv} - (f_{uv})^2 \) at critical points helps to determine the nature of those points: whether they are relative minima, maxima, or saddle points.

Relative Extremum

Relative extrema refer to the points at which a function takes on either a relative minimum or maximum value. At these points, the function value is higher or lower than at nearby points.
Using the Hessian Matrix and its determinant, we can determine whether critical points are relative extrema:

• If the Hessian determinant \( H > 0 \) and \( f_{uu} > 0 \), the point is a relative minimum.
• If \( H > 0 \) and \( f_{uu} < 0 \), the point is a relative maximum.

In our example:

• For critical point \((2, 2)\), \( H = 108 \) and \( f_{uu} = 12 \), leading to a relative minimum.

Relative extrema help us understand the function's optimization and the areas where changes in variables yield the most significant effect.

Saddle Points

Saddle points are special types of critical points where a function does not have a local extremum. Instead, the function has a mixed behavior: curves upward in one direction and downward in another, resembling a saddle.
Saddle points occur if the Hessian determinant is less than zero \( H < 0 \). This implies the critical point is neither a minimum nor a maximum.
In our function analysis:

• The critical point \((0, 0)\) shows \( H = -36 \), indicating a saddle point.
• Similarly, at \((-2, 2)\), with \( H = -180 \), we also find a saddle point.

Identifying saddle points is crucial as they illustrate the points where the function has no simple extremum behavior, showing the presence of more complex surface topographies.