Problem 2 - A First Course in Mathematical Modeling

Question

Find the local minimum value of the function $$ f(x, y)=3 x^{2}+6 x y+7 y^{2}-2 x+4 y $$

Step-by-Step Solution

Answer

The local minimum value of the function is \(\frac{25}{9}\).

Step 1: Find the Partial Derivatives

First, we need to find the partial derivatives of the function with respect to each variable. The partial derivative with respect to \(x\) is \(f_x(x, y) = \frac{\partial}{\partial x}(3x^2 + 6xy + 7y^2 - 2x + 4y) = 6x + 6y - 2\). The partial derivative with respect to \(y\) is \(f_y(x, y) = \frac{\partial}{\partial y}(3x^2 + 6xy + 7y^2 - 2x + 4y) = 6x + 14y + 4\).

Step 2: Solve for Critical Points

Set both partial derivatives equal to zero to find the critical points: \(6x + 6y - 2 = 0\) and \(6x + 14y + 4 = 0\). Solving these equations simultaneously, we first solve for \(x\) from the first equation: \(x + y = \frac{1}{3}\), and substitute \(x = \frac{1}{3} - y\) into the second equation to find \(-6y + 4 = 0\), giving us \(y = \frac{2}{3}\). Substitute back to find \(x = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}\). The critical point is \((-\frac{1}{3}, \frac{2}{3})\).

Step 3: Determine the Nature of the Critical Point

To determine whether the critical point is a minimum, we use the second derivative test. Calculate second derivatives: \(f_{xx} = 6\), \(f_{yy} = 14\), and \(f_{xy} = 6\). Evaluate the Hessian determinant at the critical point: \(D = f_{xx}f_{yy} - (f_{xy})^2 = 6 \times 14 - 6^2 = 84 - 36 = 48\). Since \(D > 0\) and \(f_{xx} > 0\), the function has a local minimum at the critical point.

Step 4: Calculate the Function Value at the Minimum

Substitute the critical point \((-\frac{1}{3}, \frac{2}{3})\) back into the original function to find the local minimum value: \(f(-\frac{1}{3}, \frac{2}{3}) = 3(-\frac{1}{3})^2 + 6(-\frac{1}{3})(\frac{2}{3}) + 7(\frac{2}{3})^2 - 2(-\frac{1}{3}) + 4(\frac{2}{3})\). Simplifying, we get \(\frac{1}{3} - \frac{4}{3} + \frac{28}{9} + \frac{2}{3} + \frac{8}{3} = \frac{25}{9}\). Thus, the local minimum value is \(\frac{25}{9}\).

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a multivariable function changes as one of the variables changes, while the others are held constant. Consider a function \( f(x, y) \) which depends on two variables: \( x \) and \( y \). To find its partial derivatives, we need to apply differentiation with respect to each variable one at a time, treating other variables as constants.

For instance, the partial derivative of \( f(x, y) \) with respect to \( x \) is denoted as \( f_x \) or \( \frac{\partial f}{\partial x} \). This tells us the rate of change of the function along the \( x \) direction. Similarly, \( f_y \) or \( \frac{\partial f}{\partial y} \) is the partial derivative with respect to \( y \), indicating the change along the \( y \) direction.

In our original exercise, we computed these derivatives to be \( f_x = 6x + 6y - 2 \) and \( f_y = 6x + 14y + 4 \). These derivatives are essential in finding critical points and understanding how the function behaves near these points.

Critical Points

Once we know the partial derivatives of a function, we can find its critical points. Critical points occur where the partial derivatives equal zero or where they do not exist. These points are crucial because they may indicate where functions reach a local minimum, maximum, or saddle point.

To locate critical points, we set \( f_x \) and \( f_y \) to zero and solve the resulting system of equations. In our example, the equations were:

• \( 6x + 6y - 2 = 0 \)
• \( 6x + 14y + 4 = 0 \)

Solving these, we find a critical point at \( \left(-\frac{1}{3}, \frac{2}{3}\right) \).
To determine if this critical point represents a local minimum, maximum, or saddle point, we use the second derivative test, assessing the curvature of the function around this point.

Second Derivative Test

The second derivative test helps classify the nature of critical points for multivariable functions. After identifying a critical point, we compute the second partial derivatives: \( f_{xx} \), \( f_{yy} \), and \( f_{xy} \).

We then evaluate the Hessian determinant \( D \) at the critical point, defined as:
\[ D = f_{xx} f_{yy} - (f_{xy})^2 \]
This value helps us understand the behavior of the function around the critical point:

• If \( D > 0 \) and \( f_{xx} > 0 \), the function has a local minimum.
• If \( D > 0 \) and \( f_{xx} < 0 \), the function has a local maximum.
• If \( D < 0 \), the function has a saddle point.
• If \( D = 0 \), the test is inconclusive.

In the original problem, we found \( D = 48 \) and \( f_{xx} = 6 \). Since \( D > 0 \) and \( f_{xx} > 0 \), the critical point \( \left(-\frac{1}{3}, \frac{2}{3}\right) \) is indeed a local minimum. This systematic approach gives us confidence in classifying the point's properties accurately.