Problem 150 - Advanced Calculus

Question

Let \(\mathrm{f}\) be the function given by \(\mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{x}^{2}+\mathrm{y}^{3}-3 \mathrm{xy}\). Find the critical points of \(\mathrm{f}(\mathrm{x}, \mathrm{y})\). Then determine whether each critical point is a relative maximum, relative minimum or saddle point of \(\mathrm{f}(\mathrm{x}, \mathrm{y})\).

Step-by-Step Solution

Answer

The critical points of the function \(f(x, y) = x^2 + y^3 - 3xy\) are (0, 0) and \((\frac{3}{4}, \frac{1}{2})\). The point (0, 0) is a saddle point, while the point \((\frac{3}{4}, \frac{1}{2})\) is a relative minimum.

Step 1: Calculate the partial derivatives

First, we need to calculate the partial derivatives of the function f(x, y) with respect to x and y, which are given by: \( \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 + y^3 - 3xy) = 2x - 3y \) \( \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 + y^3 - 3xy) = 3y^2 - 3x \)

Step 2: Set the partial derivatives equal to 0

Now, we set these partial derivatives equal to 0 to find the critical points: \(2x - 3y = 0\) \(3y^2 - 3x = 0\)

Step 3: Solve the equations for the critical points

Solve these equations simultaneously to get the critical points. From the first equation, we get: \(y = \frac{2}{3}x\) Substitute this into the second equation and we get: \(3\left(\frac{2}{3}x\right)^2 - 3x = 4x^2 - 3x = 0\) \(x(4x - 3) = 0\) Hence, x = 0 or x = 3/4 When x = 0, y = 0 and when x = 3/4, y = 1/2 So the critical points are (0, 0) and (3/4, 1/2).

Step 4: Determine the type of each critical point

To classify the critical points, we use the second partial derivative test. We calculate the second partial derivatives and the determinant D: \(D(x, y) = \left(\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x\partial y}\right)^2\right)\) Calculate the second partial derivatives: \(\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(2x - 3y) = 2\) \(\frac{\partial^2 f}{\partial x\partial y} = \frac{\partial}{\partial y}(2x - 3y) = -3\) \(\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}(3y^2 - 3x) = 6y\) Calculating the determinant D: \(D(x, y) = (2)(6y) - (-3)^2 = 12y - 9\) Now we evaluate the determinant D at each critical point: 1. At (0, 0): \(D(0, 0) = 12(0) - 9 = -9 < 0\) Since D < 0, (0, 0) is a saddle point. 2. At (3/4, 1/2): \(D\left(\frac{3}{4}, \frac{1}{2}\right) = 12\left(\frac{1}{2}\right) - 9 = 3 > 0\) Since D > 0 and \(\frac{\partial^2 f}{\partial x^2} > 0\), (\(3/4, 1/2\)) is a relative minimum. In conclusion, the critical point (0, 0) is a saddle point, and the critical point \((\frac{3}{4}, \frac{1}{2})\) is a relative minimum of the function \(f(x, y) = x^2 + y^3 - 3xy\).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially when dealing with functions of multiple variables. When we have a function \( f(x, y) \), the partial derivative with respect to \( x \) measures how \( f \) changes as \( x \) changes, while keeping \( y \) constant. Similarly, the partial derivative with respect to \( y \) shows how \( f \) changes as \( y \) changes, with \( x \) held constant. For the given function \( f(x, y) = x^2 + y^3 - 3xy \):

• The partial derivative with respect to \( x \) is \( 2x - 3y \).
• The partial derivative with respect to \( y \) is \( 3y^2 - 3x \).

To find critical points, we set these partial derivatives to zero, because critical points occur where the rate of change in any direction is zero, suggesting a local extremum or saddle point might exist. Calculating partial derivatives provides insight into the slopes of our multi-variable function at any given point.

Second Partial Derivative Test

The second partial derivative test helps us categorize critical points as relative maxima, relative minima, or saddle points. After determining the critical points by solving the system of equations obtained from the first derivatives:- We compute the second partial derivatives:

• \( \frac{\partial^2 f}{\partial x^2} = 2 \)
• \( \frac{\partial^2 f}{\partial x\partial y} = -3 \)
• \( \frac{\partial^2 f}{\partial y^2} = 6y \)

- The determinant \( D(x, y) = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 \) is calculated for each critical point.Checking this determinant will tell us:

• If \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} > 0 \), then we have a relative minimum.
• If \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), then it's a relative maximum.
• If \( D < 0 \), the point is a saddle point.

Saddle Point

A saddle point in a multivariable function is a point where the function does not have a local maximum or minimum. Instead, it is a critical point where the surface curves upwards in one direction and downwards in another, resembling a saddle shape. In our analysis for \( f(x, y) = x^2 + y^3 - 3xy \), after finding the critical points, we use the second partial derivative test. At the critical point \( (0, 0) \):

• The determinant \( D(0,0) = -9 \), which is less than 0.

This negative value indicates that the critical point is a saddle point, confirming the nature of \( (0,0) \) as neither a minimum nor a maximum, but an interesting flat point where different directional derivatives exhibit different behaviors.

Relative Minimum

A relative minimum of a multivariable function is a point where the function value is lower than all nearby points. It indicates a "valley" in the surface of the graph. In the provided problem, after finding and analyzing the critical point \( \left(\frac{3}{4}, \frac{1}{2}\right) \), we find the following:

• The determinant \( D\left(\frac{3}{4}, \frac{1}{2}\right) = 3 \), which is greater than 0.
• \( \frac{\partial^2 f}{\partial x^2} = 2 \) is positive.

These conditions confirm that this point is a relative minimum. This means that moving in any small direction away from this point will increase the value of the function, indicating a dip or valley at \( \left(\frac{3}{4}, \frac{1}{2}\right) \). Understanding relative minima is key in optimization problems where identifying least values is essential.