Answered: find the local maximum and minimum…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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find the local maximum and minimum values and saddle point(s) of the function. if you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. (enter your answers in a comma separated list. If an answer does not exist, enter DNE.) f(x,y) = x^4 + y^4 -4xy + 4

Expert Solution

Step 1

Step:-1

To find maximum, minimum and saddle point of a function of multi-variable, first we find critical point and then use a test to check these points are maximum, minimum or saddle points.

Given that $f (x, y) = x^{4} + y^{4} - 4 x y + 4$

now, differentiate f(x,y) with respect to x and y, we get

$f_{x} = 4 x^{3} + 0 - 4 y + 0 = 4 x^{3} - 4 y f_{x} = 4 (x^{3} - y) f_{y} = 0 + 4 y^{3} - 4 x + 0 = 4 y^{3} - 4 x f_{y} = 4 (y^{3} - x)$

Step:-2

For critical points put $f_{x} = 0 a n d f_{y} = 0$ . So,

$f_{x} = 0 \Rightarrow 4 (x^{3} - y) = 0 \Rightarrow (x^{3} - y) = 0 \Rightarrow y = x^{3} - - - (1) a n d f_{y} = 0 \Rightarrow 4 (y^{3} - x) = 0 \Rightarrow (y^{3} - x) = 0 \Rightarrow x = y^{3} \Rightarrow y = x^{\frac{1}{3}} - - - (2) b y (1) a n d (2), w e h a v e x^{\frac{1}{3}} = x^{3} \Rightarrow x = x^{9} \Rightarrow (x^{9} - x) = 0 \Rightarrow x (x^{8} - 1) = 0 N o t e : - (x^{8} - 1) = ({(x^{4})}^{2} - 1) = (x^{4} - 1) (x^{4} + 1) = (x^{2} - 1) (x^{2} + 1) (x^{4} + 1) = (x - 1) (x + 1) (x^{2} + 1) (x^{4} + 1) \Rightarrow x (x^{8} - 1) = 0 \Rightarrow x (x - 1) (x + 1) (x^{2} + 1) (x^{4} + 1) = 0 \Rightarrow x = 0, 1, - 1$

Note:- we take real values of x only, not complex values.

by equation (1), we have

$y = x^{3} f o r x = 0 \Rightarrow y = 0 f o r x = 1 \Rightarrow y = 1 f o r x = - 1 \Rightarrow y = - 1 S o, t h e c r i t i c a l p o i n t s a r e (0, 0), (- 1, - 1), (1, 1)$

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