Answered: 8 x - y a(x, y)

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter1: Fundamental Concepts Of Algebra

Section1.2: Exponents And Radicals

Problem 29E

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Question

Find the critical points of the following multivariable function, and using the Second Derivative Test to find the maxima and minima of the equation below:

Expert Solution

Step 1

Given :- The given multivariable function is $a (x, y) = \frac{8 x - y}{e^{x^{2} + y^{2}}}$ .

To determine :- The critical points of the multivariable function and using the second derivative test to find the maxima and minima of the equation.

Step 2

Solution :-

The given multivariable function is $a (x, y) = \frac{8 x - y}{e^{x^{2} + y^{2}}}$ .

Now, differentiating the function,

$a = \frac{8 x}{e^{x^{2} + y^{2}}} - \frac{y}{e^{x^{2} + y^{2}}} a_{x} = \frac{8 [e^{x^{2} + y^{2}} . 1 - x . e^{x^{2} + y^{2}} . 2 x]}{{(e^{x^{2} + y^{2}})}^{2}} a_{x} = \frac{8 [1 - 2 x^{2}]}{(e^{x^{2} + y^{2}})} a_{y} = \frac{- [e^{x^{2} + y^{2}} . 1 - y . e^{x^{2} + y^{2}} . 2 y]}{{(e^{x^{2} + y^{2}})}^{2}} a_{y} = \frac{- [1 - 2 y^{2}]}{(e^{x^{2} + y^{2}})}$

For critical points, $a_{x} = 0, a_{y} = 0$

$a_{x} = \frac{8 [1 - 2 x^{2}]}{{(e^{x^{2} + y^{2}})}^{2}} = 0 1 - 2 x^{2} = 0 2 x^{2} = 1 x = \pm \frac{1}{\sqrt{2}} a_{y} = \frac{- [1 - 2 y^{2}]}{{(e^{x^{2} + y^{2}})}^{2}} = 0 1 - 2 y^{2} = 0 2 y^{2} = 1 y = \pm \frac{1}{\sqrt{2}}$

The critical points are $(\pm \frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}})$ .

$a_{x} = \frac{8 (1 - 2 x^{2})}{e^{x^{2} + y^{2}}} a_{x x} = \frac{8 (- 6 x + 4 x^{3})}{e^{x^{2} + y^{2}}} a_{y} = \frac{- (1 - 2 y^{2})}{e^{x^{2} + y^{2}}} a_{y y} = \frac{- (6 y + 4 y^{3})}{e^{x^{2} + y^{2}}}$

$a_{x y} = 8 (1 - 2 x^{2}) \frac{- 2 y}{{(e^{x^{2} + y^{2}})}^{2}}$

At point $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ ,

$a_{x x} = \frac{8 (- 6 x + 4 x^{3})}{e^{x^{2} + y^{2}}} = - \frac{32}{\sqrt{2} e} a_{y y} = \frac{- (- 6 y + 4 y^{3})}{e^{x^{2} + y^{2}}} = \frac{4}{\sqrt{2} e}$

$a_{x y} = 8 (1 - 2 x^{2}) \frac{- 2 y}{{(e^{x^{2} + y^{2}})}^{2}} = 0$

Hence,

$= a_{x x} a_{y y} - a_{x y} = (\frac{- 32}{\sqrt{2} e}) (\frac{4}{\sqrt{2} e}) - 0 = \frac{- 64}{e^{2}} = - 8.6615$

Now, thus this value is less than 0 then at this point there is neither maximum nor minimum and have a saddle point at $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ .

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