Problem 8 - 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. \(f(x, y)=2 x^{3}+y^{2}-6 x^{2}-4 y+12 x-2\)

Step-by-Step Solution

Answer

The critical point of the given function \(f(x, y) = 2x^3 + y^2 - 6x^2 - 4y + 12x - 2\) is \((1, 2)\). However, since the second derivative test is inconclusive at this point, we cannot determine the nature of the critical point and thus cannot determine the relative extrema of the function.

Step 1: Find the first partial derivatives

To find the first partial derivatives of the function \(f(x, y)\) with respect to x and y, we'll differentiate the function partially. The first partial derivative of \(f(x, y)\) with respect to x is: \(f_x(x, y) = \frac{\partial f}{\partial x} = 6x^2 - 12x + 12\) The first partial derivative of \(f(x, y)\) with respect to y is: \(f_y(x, y) = \frac{\partial f}{\partial y} = 2y - 4\)

Step 2: Find the critical points

To find the critical points, we need to solve the system of equations formed by setting both first partial derivatives equal to zero: \(f_x(x, y) = 6x^2 - 12x + 12 = 0\) \(f_y(x, y) = 2y - 4 = 0\) Solving the second equation first, we get: \(y = 2\) Plugging the value of y in the first equation: \(6x^2 - 12x + 12 = 0\) Dividing by 6, we get: \(x^2 - 2x + 2 = 0\) By solving the above quadratic equation, we get two critical points: \(x = 1\pm\sqrt{1}\) So, the critical points are: \((1, 2)\), \((1, 2)\)

Step 3: Second Derivative Test

Now we'll find the second partial derivatives of the function and use them to determine the nature of the critical points: The second partial derivative with respect to x: \(f_{xx}(x, y) = \frac{\partial^2 f}{\partial x^2} = 12x-12\) The second partial derivative with respect to y: \(f_{yy}(x, y) = \frac{\partial^2 f}{\partial y^2} = 2\) The mixed second derivative with respect to x and y: \(f_{xy}(x, y) = \frac{\partial^2 f}{\partial x\partial y} = 0\) Now we'll calculate the discriminant D using these second partial derivatives: \(D(x, y) = f_{xx}(x, y)f_{yy}(x, y) - f_{xy}(x, y)^2\) At point (1, 2): \(D(1, 2) = (12 - 12)(2) - 0^2 = 0\) Since D is equal to 0, the second derivative test is inconclusive at the critical point (1, 2).

Step 4: Determine the relative extrema

As the second derivative test is inconclusive, we cannot determine the nature of the critical point and thus cannot determine the relative extrema of the function, based on the given critical points.

Key Concepts

Second Derivative Test

The second derivative test is a powerful tool in multivariable calculus. It's used to determine the nature of critical points. A critical point occurs where the first partial derivatives are zero. In this context, we want to classify these critical points as either local maxima, local minima, or saddle points.

To do this, we calculate the second partial derivatives of the function. These include:

• The second partial derivative with respect to x: \( f_{xx} \)
• The second partial derivative with respect to y: \( f_{yy} \)
• The mixed partial derivative: \( f_{xy} \)

The key formula here is the discriminant, \( D \), which is defined as:\[ D(x, y) = f_{xx}(x, y) f_{yy}(x, y) - (f_{xy}(x, y))^2 \]

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

Understanding this test is crucial because it allows us to analyze critical points beyond simply identifying them, helping us determine the behavior of the function at these points.

Partial Derivatives

Partial derivatives are used to examine how a multivariate function changes as we vary one variable while keeping others constant. In this exercise, we deal with a function of two variables, \( f(x, y) = 2x^3 + y^2 - 6x^2 - 4y + 12x - 2 \).

When finding the partial derivative with respect to x, denoted as \( f_x(x, y) \), we treat \( y \) as a constant. This results in the expression:\[ f_x(x, y) = 6x^2 - 12x + 12 \]
Similarly, when computing the partial derivative with respect to y, denoted as \( f_y(x, y) \), consider \( x \) as constant, giving us:\[ f_y(x, y) = 2y - 4 \]

The critical points are found where these derivatives equal zero, providing insights into where the function's rate of change in both directions is zero. These become potential points for local maxima, minima, or saddle points. Knowledge of partial derivatives is vital for anyone looking to understand the behavior of functions involving more than one variable.

Relative Extrema

Relative extrema are points where the function reaches a local maximum or minimum within a specific region. They are fundamentally important in optimization and finding optimal solutions.

Once we identify critical points through the first partial derivatives, the second derivative test assesses if these points are relative extrema. The function \( f(x, y) \) used here introduces critical points based on the system of derivative equations, where both partial derivatives equal zero.

In this particular exercise, the second derivative test is inconclusive because the determinant \( D \) equaled zero at the critical point. This means the nature of extremum isn't easily classified.

In such cases, additional analysis or graphical methods might be necessary to determine whether these points are indeed points of local maxima, minima, or saddle points. Understanding relative extrema is vital in real-world applications, including economics, engineering, and physics, where optimizing function values is often required. This understanding allows us to make informed decisions based on function behavior.