Short Answer

$Extrema : Find the local maxima, local minima, and saddle points of the given functions f (x, y) = x^{3} + y^{3} - 6 x^{2} + 3 y^{2} - 4 .$

$Critical point are (0, 0), (0, - 2), (4, 0), (4, - 2)$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given

$f (x, y) = x^{3} + y^{3} - 6 x^{2} + 3 y^{2} - 4 .$

Step 2. Finding critical point .

$Given function is f (x, y) = x^{3} + y^{3} - 6 x^{2} + 3 y^{2} - 4, which is a polynomial and hence differentiable . The gradient of this function is \nabla f (x, y) = \frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j = (3 x^{2} - 12 x) i + (3 y^{2} + 6 y) j At the critical points the gradient of a function vanishes, that is \nabla f (x, y) = 0 . from the above the critical points are given by 3 x^{2} - 12 x = 0 . . . (1) and 3 y^{2} + 6 y = 0 . . . . . (2) from (1) 3 x^{2} - 12 x = 0, the roots of which are x = 0, 4 from (1) 3 y^{2} + 6 y = 0, the roots of which are y = 0, - 2 Solving these, critical point as (0, 0), (0, - 2), (4, 0), (4, - 2) .$

Step 3. Finding maxima and minima .

$Now the second order derivatives of the given function are \frac{\partial^{2} f}{\partial x^{2}} = 6 x - 12, \frac{\partial^{2} f}{\partial y^{2}} = 6 y + 6, \frac{\partial^{2} f}{\partial y^{} \partial x} = - 0 . Hence the discriminate of the given function is H_{f} (x, y) = \frac{\partial^{2} f}{\partial x^{2}} \frac{\partial^{2} f}{\partial y^{2}} - {(\frac{\partial^{2} f}{\partial y^{} \partial x})}^{2} = (6 x - 12) (6 + 6 y) - {(0)}^{2} = 36 (xy + x - 2 y - 2) . At (0, 0), H_{f} (0, 0) = - 72 < 0 So the critical point (0, 0) is a saddle point . At (0, - 2), H_{f} (0, - 2) = 72 > 0 also at (0, - 2) \frac{\partial^{2} f}{\partial x^{2}} = - 12 < 0, \frac{\partial^{2} f}{\partial y^{2}} = - 6 < 0 Therefore the critical point is (0, - 2) is local maxima . At (4, 0), H_{f} (4, 0) = 72 > 0 also at (4, 0) \frac{\partial^{2} f}{\partial x^{2}} = 12 > 0, \frac{\partial^{2} f}{\partial y^{2}} = 6 > 0 Therefore the critical point is (4, 0) is local minima . At (4, - 2), H_{f} (4, - 2) = - 72 < 0 So the critical point (4, - 2) is a saddle point .$

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Calculus

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Short Answer

$Extrema : Find the local maxima, local minima, and saddle points of the given functions f (x, y) = x^{3} + y^{3} - 6 x^{2} + 3 y^{2} - 4 .$

Step by Step Solution

Step 1. Given

Step 2. Finding critical point .

Step 3. Finding maxima and minima .

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Calculus

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Extrema: Find the local maxima, local minima, and saddle points of the given functionsf(x, y) = x3+ y3 − 6x2 + 3y2-4.

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Step 1. Given

Step 2. Finding critical point .

Step 3. Finding maxima and minima .

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$Extrema : Find the local maxima, local minima, and saddle points of the given functions f (x, y) = x^{3} + y^{3} - 6 x^{2} + 3 y^{2} - 4 .$