Problem 6 - 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. \(f(x, y)=x^{2}-4 x y+2 y^{2}+4 x+8 y-1\)

Step-by-Step Solution

Answer

The critical point of the function \(f(x, y) = x^2 - 4xy + 2y^2 + 4x + 8y - 1\) is (6, 4). Using the second derivative test, we found that this point is a saddle point.

Step 1: Partial Derivative with respect to x

To find the partial derivative of the given function with respect to x, differentiate it with respect to x while treating y as a constant: \[ \frac{\partial f}{\partial x} = 2x - 4y + 4 \]

Step 2: Partial Derivative with respect to y

Similarly, differentiate the function with respect to y, treating x as a constant: \[ \frac{\partial f}{\partial y} = -4x + 4y + 8 \] Now, set both partial derivatives equal to zero and solve the system of equations: \[ \begin{cases} 2x - 4y + 4 = 0\\ -4x + 4y + 8 = 0 \end{cases} \]

Step 3: Solving for Critical Points

From equation 1, we have 2x = 4y - 4. Since both sides can be divided by 2, we can rewrite the equation as x = 2y - 2. Now, substitute the expression for x in equation 2: \[-4(2y - 2) + 4y + 8 = 0\] Simplifying the equation, we get: \[-8y + 8 + 4y + 8 = 0\] \[-4y + 16 = 0\] Divide both sides by 4: \[-y + 4 = 0\] Therefore, y = 4. Then substituting y back into the expression for x: \[x = 2(4) - 2 = 6\] Thus, (6, 4) is the critical point of the function. Step 2: Find the second partial derivatives

Step 4: Second Partial Derivatives

Find the second partial derivatives of the function with respect to x and y: \[ \frac{\partial^2 f}{\partial x^2} = 2 \] \[ \frac{\partial^2 f}{\partial x \partial y} = -4 \] \[ \frac{\partial^2 f}{\partial y^2} = 4 \] Step 3: Classify the critical points using the second derivative test

Step 5: Second Derivative Test

Use the second derivative test to classify the critical points of the function: The second derivative test formula is: \[D(x, y) = \left(\frac{\partial^2 f}{\partial x^2}\right) \left(\frac{\partial^2 f}{\partial y^2}\right) - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2\] Calculate D(6, 4): \[D(6, 4) = (2)(4) - (-4)^2 = 8 - 16 = -8\] Since D(6, 4) < 0, the critical point (6, 4) is a saddle point. Finally, we can conclude that the function \(f(x, y) = x^2 - 4xy + 2y^2 + 4x + 8y - 1\) has a saddle point at (6, 4).

Key Concepts

Partial Derivatives

When we study multivariable functions like f(x, y), we use partial derivatives to measure how the function changes as one variable varies, keeping all other variables constant. Just as a derivative in single-variable calculus represents the slope of the tangent line to a curve, a partial derivative represents the slope of the tangent plane to a surface along an axis.

For example, to find the partial derivative of f(x, y) = x^2 - 4xy + 2y^2 + 4x + 8y - 1 with respect to x, we treat y as a constant and differentiate with respect to x. This process yields ∂f/∂x = 2x - 4y + 4, which gives us information about the function's rate of change in the direction of x. Similarly, the partial derivative with respect to y, ∂f/∂y = -4x + 4y + 8, tells us the rate of change in the direction of y. By setting these derivatives to zero, we find the critical points, locations where the function's rate of change switches direction or is momentarily stationary.

Saddle Point

A saddle point is a type of critical point in multivariable calculus analogous to a high point on a horse's saddle where the surface curves upward in one direction and downward in another. Mathematically, at a saddle point, the function does not have a maximum or minimum but has a zero gradient.

In our example, after computing the partial derivatives and setting them to zero, we find that the function f(x, y) has a saddle point at (6, 4). The determination of a saddle point is often made using the second derivative test for functions of two variables. If the value D from this test is negative, as in our case where D(6, 4) = -8, it indicates that the critical point is a saddle point.

Second Derivative Test

The Second Derivative Test is a handy tool for determining the nature of a critical point found in the domain of a function of two variables. Once we've located a critical point by finding where the partial derivatives equal zero, we apply this test to identify whether the point is a local maximum, a local minimum, or a saddle point.

The test involves computing the second partial derivatives of the function and substituting the coordinates of the critical point into the formula D(x, y) = (∂²f/∂x²)(∂²f/∂y²) - (∂²f/∂x∂y)². If D is positive, and ∂²f/∂x² is positive, then we have a relative minimum. If D is positive and ∂²f/∂x² is negative, it is a relative maximum. A negative D signifies a saddle point, indicating that the surface is curved in opposite directions along the x and y axes.

Relative Extrema

Relative extrema, also known as local maxima or minima, are points in the domain of a function where the function has smaller or larger values than at nearby points. Think of them as local peaks or valleys in the function's graph. In multivariable functions, these points are critical for understanding the topography of surfaces.

To locate relative extrema, we look for critical points by setting the partial derivatives to zero and solving for the variables. However, not all critical points are relative extrema; some could be saddle points. The Second Derivative Test helps us distinguish between these possibilities. In our example, the second derivative test showed that the critical point (6, 4) was a saddle point, hence the function does not exhibit any relative extrema at this point.