Problem 20 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. \(f(x, y)=x y+\ln x+2 y^{2}\)

Step-by-Step Solution

Answer

The critical point of the function \(f(x, y) = xy + \ln x + 2y^2\) is (-2, 0.5). By using the second derivative test, we found that the determinant of the Hessian matrix is negative, indicating that this critical point is a saddle point. Therefore, there are no relative extrema (local maxima or minima) for this function.

Step 1: Find the first partial derivatives

To find the first partial derivatives, we need to differentiate the function f(x, y) with respect to both x and y. The first partial derivative with respect to x: \(\frac{\partial}{\partial x}(x y + \ln x + 2y^2) = y + \frac{1}{x}\) The first partial derivative with respect to y: \(\frac{\partial}{\partial y}(x y + \ln x + 2y^2) = x + 4y\)

Step 2: Find the critical points

Critical points occur when the gradient of the function is equal to the zero vector, that is, both partial derivatives are equal to zero. So we need to solve the following system of equations: \(y + \frac{1}{x} = 0\) \(x + 4y = 0\) From the second equation, we can express x in terms of y: \(x = -4y\) Now substitute this into the first equation: \(-4y^2 + \frac{1}{-4y} = 0\) This equation can be simplified further by multiplying both sides by -4y: \(16y^3 - 1 = 0\) Now we can find the value of y by solving this cubic equation: \(y^3 = \frac{1}{16}\) \(y = \frac{1}{2}\) Now substitute the value of y back into the equation x = -4y: \(x = -4(\frac{1}{2}) = -2\) So the critical point is (-2, 0.5)

Step 3: Compute the second partial derivatives

Now we need to find the second partial derivatives: \(\frac{\partial^2}{\partial x^2}(f) = -\frac{1}{x^2}\) \(\frac{\partial^2}{\partial x \partial y}(f) = 1\) \(\frac{\partial^2}{\partial y^2}(f) = 4\)

Step 4: Calculate the determinant of the Hessian matrix

To classify the critical point, we can use the second derivative test which requires us to compute the determinant of the Hessian matrix. The determinant of the Hessian matrix is given by D = \(\left(\frac{\partial^2}{\partial x^2}\right)\left(\frac{\partial^2}{\partial y^2}\right)-\left(\frac{\partial^2}{\partial x \partial y}\right)^2 \) Now plug in the values of the second partial derivatives at the critical point (-2, 0.5): \(D = \left(-\frac{1}{(-2)^2}\right)(4) - (1)^2\) \(D = \frac{1}{4}(4) - 1 = -\frac{1}{2}\)

Step 5: Use the second derivative test to classify the critical point

Since the determinant D < 0, the critical point (-2, 0.5) is a saddle point.

Step 6: Determine the relative extrema of the function

In this case, since the only critical point is a saddle point, there are no relative extrema (local maxima or minima) for the function f(x,y).

Key Concepts

Partial Derivatives

To start finding critical points, partial derivatives are essential. Partial derivatives involve differentiating a function of multiple variables with respect to one variable at a time, while keeping the others constant.

In our function, \(f(x, y) = xy + \ln x + 2y^2\), we calculated:

• \(\frac{\partial f}{\partial x} = y + \frac{1}{x}\)
• \(\frac{\partial f}{\partial y} = x + 4y\)

Setting these derivatives equal to zero helps us find the critical points, representing where the function's slope levels out.

By solving these equations, we found the critical point \((-2, 0.5)\). Partial derivatives, hence, serve as tools to locate potential maxima, minima, or saddle points of multivariable functions.

Second Derivative Test

Once we locate the critical points, the second derivative test helps us classify them. It's based on evaluating the second derivatives.

For our function, we need to compute the mixed and second partial derivatives:

• \(\frac{\partial^2 f}{\partial x^2} = -\frac{1}{x^2}\)
• \(\frac{\partial^2 f}{\partial y^2} = 4\)
• \(\frac{\partial^2 f}{\partial x \partial y} = 1\)

With these, we can perform the second derivative test using the Hessian matrix, which reveals the curvature behavior at the critical point.

This test can conclusively show if a critical point is a local maximum, minimum, or saddle point based on the determinants of the Hessian.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives, central to understanding the behavior around critical points. For a function \(f(x, y)\), the Hessian is:

• \[H = \begin{bmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix}\]

For our function, we calculated:

• \(D = \left(-\frac{1}{4}\right) \cdot 4 - 1^2 = -\frac{1}{2}\)

Here, \(D\) is the determinant of the Hessian. The sign of \(D\) helps determine the nature of the critical point.

In this exercise, \(D < 0\) indicates a saddle point at \((-2, 0.5)\), showing the function changes concavity in different directions.

Relative Extrema

Relative extrema refer to points where a function reaches local minima or maxima. To find these, one must identify critical points and apply the second derivative test.

In cases where the Hessian determinant \(D > 0\), further checks tell us whether it’s a minimum or maximum. If the value is positive throughout, it's a local minimum.

Our function with \(D < 0\) at \((-2, 0.5)\) confirmed a saddle point—meaning neither a local max nor min, but rather a point where the surface curves upward in one direction and downward in another.

Understanding these concepts helps in analyzing real-world multivariable systems, often revealing critical insights into their behavior and equilibrium.