Problem 4 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 6 , determine the relative extrema of \(f\), if there are any. \(f(x, y)=x^{3}+y^{3}-18 x y\)

Step-by-Step Solution

Answer

The function has a saddle point at (0,0) and a local minimum at (6,6).

Step 1 - Find the First Partial Derivatives

Compute the partial derivatives of the function with respect to both variables. \(\frac{\text{∂}f}{\text{∂}x} = 3x^2 - 18y\) and \(\frac{\text{∂}f}{\text{∂}y} = 3y^2 - 18x\).

Step 2 - Set the Partial Derivatives to Zero

Set each of the first partial derivatives to zero to find the critical points: \(3x^2 - 18y = 0\) and \(3y^2 - 18x = 0\).

Step 3 - Solve the System of Equations

Solve the system of equations resulting from the zero partial derivatives: \( x^2 = 6y \) and \( y^2 = 6x \). Substitute one equation into the other to solve for \(x\) and \(y\).

Step 4 - Find the Critical Points

Substitute \( y = \frac{x^2}{6} \) into \( y^2 = 6x \): \( \frac{x^4}{36} = 6x \). Hence, \( x = 0 \) and \( x^3 = 216 \rightarrow x = 6 \). Corresponding \( y \) values are \( y = 0 \) and \( y = 6 \), giving critical points (0, 0) and (6, 6).

Step 5 - Use the Second Derivative Test

Employ the second derivative test to determine the nature of each critical point: Compute \( f_{xx}(x, y) = 6x \), \( f_{yy}(x, y) = 6y \), and \( f_{xy}(x, y) = -18 \). Then use the determinant of the Hessian matrix: \( D = f_{xx} f_{yy} - (f_{xy})^2 \).

Step 6 - Evaluate the Determinant at Critical Points

Evaluate \( D \) at (0,0) and (6,6): At (0,0): \( D(0,0) = 0 \times 0 - (-18)^2 = -324 \), indicating a saddle point. At (6,6): \( D(6,6) = 36 \times 36 - (-18)^2 = 1296-324 = 972 \). Since \( f_{xx}(6, 6) = 36 \(> 0\) \), (6,6) is a local minimum.

Key Concepts

Partial Derivatives

Partial derivatives are a way to understand how a multi-variable function changes as we vary one variable, keeping the other variables constant. Consider our function:
\( f(x, y) = x^3 + y^3 - 18xy \)
To find the partial derivative with respect to x, we treat y as a constant and differentiate:
\( \frac{\partial f}{\partial x} = 3x^2 - 18y \)
Similarly, for the partial derivative with respect to y, treat x as a constant:
\( \frac{\partial f}{\partial y} = 3y^2 - 18x \)
These partial derivatives tell us the rate of change of the function in the direction of x and y, respectively. They're crucial for finding the critical points.

Critical Points

Critical points occur where the first partial derivatives of the function are equal to zero. They represent points where the function's rate of change is zero in all directions. To find them:
- Set the partial derivatives equal to zero: \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 0 \)
This will gives us:
\( 3x^2 - 18y = 0 \) and \( 3y^2 - 18x = 0 \)
Solving this system of equations, we get critical points \( (0,0) \) and \( (6,6) \). These points are where potential relative extrema (maximum or minimum values) could occur.

Second Derivative Test

The Second Derivative Test helps us determine the nature of the critical points (whether they are minima, maxima, or saddle points). We need to compute the second partial derivatives and the Hessian determinant.
First, find the second partial derivatives:
\( f_{xx} = 6x \)
\( f_{yy} = 6y \)
\( f_{xy} = -18 \)
The Hessian determinant is defined as: \[ D = f_{xx}f_{yy} - (f_{xy})^2 \]
Evaluate D at the critical points:
- At (0,0): \[ D(0,0) = 6(0) \times 6(0) - (-18)^2 = -324 \]
This indicates a saddle point.
- At (6,6): \[ D(6,6) = 6(6) \times 6(6) - (-18)^2 = 972 \]
Since \( f_{xx}(6, 6) > 0 \), (6,6) is a local minimum.
Use these results to understand the function's behavior around critical points accurately.