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Q. 37

Expert-verified

Calculus

Found in: Page 989

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Extrema: Find the local maxima, local minima, and saddle points of the given functions.
$f (x, y) = 2 x^{2} + y^{2} + y + 5 .$

$Critical point as (0, - \frac{1}{2}) and the local minima point for the function is (0, - \frac{1}{2})$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1 . Given

$f (x, y) = 2 x^{2} + y^{2} + y + 5 .$

Step 2. Finding saddle points.

$Given function is f (x, y) = 2 x ² + y ² + y + 5, which is a polynomial and hence differentiable . The gradient of this function is \nabla f (x, y) = \frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j = 4 xi + (2 y + 1) j At the critical points the gradient of a function vanishes, that is \nabla f (x, y) = 0 . from the above the critical points are given by 4 x = 0 and 2 y + 1 = 0 . Solving these two gives one critical point as (0, - \frac{1}{2}) .$

Step 3. Finding maxima and minima .

$Now the second order derivatives of the given function are \frac{\partial^{2} f}{\partial x^{2}} = 4, \frac{\partial^{2} f}{\partial y^{2}} = 2, \frac{\partial^{2} f}{\partial y^{} \partial x} = 0 . Hence the discriminate of the given function is H_{f} (x, y) = \frac{\partial^{2} f}{\partial x^{2}} \frac{\partial^{2} f}{\partial y^{2}} - {(\frac{\partial^{2} f}{\partial y^{} \partial x})}^{2} = 4 (2) - 0 = 8 . Since, H_{f} (x, y) = 8 > 0 and \frac{\partial^{2} f}{\partial x^{2}} = 4 > 0, \frac{\partial^{2} f}{\partial y^{2}} = 2 > 0 Therefore the local minima point for the function is (0, - \frac{1}{2})$

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