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Chapter 13: Problem 12

Examine the function for relative extrema and saddle points. $$ f(x, y)=-3 x^{2}-2 y^{2}+3 x-4 y+5 $$

Short Answer

Expert verified
The function \( f(x, y)=-3 x^{2}-2 y^{2}+3 x-4 y+5 \) has a relative maximum at the point (0.5, -1).

Step by step solution

01

Find Partial Derivatives

The first step is to find the first-order partial derivatives of \( f \) with respect to \( x \) and \( y \). These are calculated as follows:\n\nThe derivative of \( f \) with respect to \( x \) is \( f_{x}(x, y) = -6x + 3 \).\n\nThe derivative of \( f \) with respect to \( y \) is \( f_{y}(x, y) = -4y - 4 \).
02

Find Critical Points

Next, we will find the critical points by setting these first-order partial derivatives equal to zero and solving for \( x \) and \( y \). This gives us:\n\n\( -6x + 3 = 0 \) \n\n\( -4y - 4 = 0 \) \n\nSolving these equations, we find the critical point (0.5, -1).
03

Classification of Critical Points

To classify the critical points, we first find the second-order partial derivatives:\n\nThe derivative of \( f_{x} \) with respect to \( x \) is \( f_{xx}(x, y) = -6 \).\nThe derivative of \( f_{y} \) with respect to \( y \) is \( f_{yy}(x, y) = -4 \).\n\nThe mixed derivative of \( f_{x} \) with respect to \( y \) is \( f_{xy}(x, y) = 0 \).\n\nWe now consider the determinant of the Hessian matrix \((H)\) at the critical point. The Hessian matrix is given by:\n\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix}\]\n\nSo, the determinant is: \( D = f_{xx}f_{yy} - f_{xy}^{2} = -6*-4 - 0 = 24\).\n\nSince \( D>0 \) and \( f_{xx}<0 \), the critical point (0.5, -1) is a relative maximum.

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