Problem 9 - College Algebra and Calculus: An Applied Approach

Question

Examine the function for relative extrema and saddle points. $$ f(x, y)=x^{2}+6 x y+10 y^{2}-4 y+4 $$

Step-by-Step Solution

Answer

The function \(f(x, y)=x^{2}+6 x y+10 y^{2}-4 y+4\) has a relative minimum at the point (0.4, -0.2). There are no relative maxima or saddle points.

Step 1: Calculating the First Partial Derivatives

Calculate the partial derivatives of the function with respect to \(x\) and \(y\). The first order partial derivatives are defined as follows: \[ f_{x}=\frac{\partial f}{\partial x}=2x+6y \] And \[ f_{y}=\frac{\partial f}{\partial y}=6x+20y-4 \]

Step 2: Finding the Critical Points

The critical points can be found by setting \(f_x = 0\) and \(f_y = 0\) and then solving the resulting system of equations. This gives us the system of equations \[2x+6y = 0\] and \[6x+20y-4 = 0\]. Solving this system gives: \(x = 0.4\) and \(y = -0.2\).

Step 3: Calculating the Second Partial Derivatives

Evaluate the second order partial derivatives: \[f_{xx} = \frac{\partial^{2} f}{\partial x^2}=2\]\[f_{yy} = \frac{\partial^{2} f}{\partial y^2}=20\]\[f_{xy} = \frac{\partial^{2} f}{\partial x \partial y} = 6\]

Step 4: Second Derivative Test

Use second derivative test to categorize the critical points. The discriminant \(D\) is given by following formula: \[D = f_{xx}f_{yy} - {f_{xy}}^2\] Substituting the values, we get \(D = (2)(20) - (6)^2 = 24\). Since \(D\) is positive, and \(f_{xx}\) is also positive, the function has a relative minimum at the critical point (0.4, -0.2).