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Chapter 13: Problem 9

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

Short Answer

Expert verified
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 by step solution

01

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 \[ f_{y}=\frac{\partial f}{\partial y}=6x+20y-4 \]
02

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 \(y = -0.2\).
03

Calculating the Second Partial Derivatives

Evaluate the second order partial derivatives: \[f_{xx} = \frac{\partial^{2} f}{\partial x^2}=2\]\[f_{yy} = \frac{\partial^{2} f}{\partial y^2}=20\]\[f_{xy} = \frac{\partial^{2} f}{\partial x \partial y} = 6\]
04

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).

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