topics_15_7_Min_and_Max_Values_Part_III_Extremal_Problems

# topics_15_7_Min_and_Max_Values_Part_III_Extremal_Problems -...

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Unformatted text preview: 15.7. Max and Min Values. Part III. Extremal Problems. Problem 40/ page 968. Find the point on the plane x- y + z = 4 that is closest to the point (1 , 2 , 3). Solution. min d = p ( x- 1) 2 + ( y- 2) 2 + ( z- 3) 2 subject to x- y + z = 4 . Now, instead of looking for min d = p ( x- 1) 2 + ( y- 2) 2 + ( z- 3) 2 it is computationally more simple to look for min d 2 = ( x- 1) 2 + ( y- 2) 2 + ( z- 3) 2 . Hence, we shall solve min d 2 = ( x- 1) 2 + ( y- 2) 2 + ( z- 3) 2 subject to x- y + z = 4 . Then, z = 4- x + y and we have to minimize f ( x,y ) = ( x- 1) 2 + ( y- 2) 2 + (1- x + y ) 2 ,-∞ < x < ∞ ,-∞ < y < ∞ ( x- 1) 2 + ( y- 2) 2 + (1- x + y ) 2 = 2 x 2 + 2 y 2 + ax + by + c for some numbers a,b,c and from here, obviously, f ( x,y ) > 2 r 2- | a || r | - | b || r | - | c | > f (0 , 0) = 6 for x 2 + y 2 ≥ r, where r sufficiently big. From here, the minimum value of f ( x,y ) on the plane R 2 is equal to the minimum value of f ( x,y ) on the domain { ( x,y ) : x 2 + y 2 ≤ r } . Then according to the Extreme....
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topics_15_7_Min_and_Max_Values_Part_III_Extremal_Problems -...

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