Spring2011Quiz6Solution

Spring2011Quiz6Solution - whose components have the largest...

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MAC2313 Quiz 6 – Solution 1. Find the extreme values of f ( x,y ) = x 2 - y 2 on the disk D = { ( x,y ) | x 2 + y 2 1 } . Solution: First find the critial points of f that lie in the domain D : f x ( x,y ) = 2 x f y ( x,y ) = - 2 y It follows that (0 , 0) is the only critical point of f and it lies in the domain D . Since f xx ( x,y ) = 2 f xy ( x,y ) = 0 f yx ( x,y ) = 0 f yy ( x,y ) = - 2 we have D ( x,y ) = - 4 < 0 for any ( x,y ). Hence, (0 , 0) is a saddle point. The boundaries of the domain are given by the functions f ( x, p 1 - x 2 ) = x 2 - (1 - x 2 ) = 2 x 2 - 1 f ( x, - p 1 - x 2 ) = x 2 - (1 - x 2 ) = 2 x 2 - 1 On each the first function the point (0 , 1) is a critical point, and on the second function the point (0 , - 1) is a critical point. The boundary points for both function are (1 , 0) and ( - 1 , 0). These 4 points are the potential absolute maximum/minimum points of f on D . Absolute Minimum (Points; Value) Absolute Maximum (Points; Value) (0 , 1) , (0 , - 1) ; - 1 (1 , 0) , ( - 1 , 0) ; 1 2. Find a vector in R 3 whose length is 16, assume the z -component is non-negative, and
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Unformatted text preview: whose components have the largest possible sum. (Find only the critical points.) Solution: Assume h x,y,z i is a vector in R 3 such that z ≥ 0 such that ||h x,y,z i|| = 16 and Sum = x + y + z . Solving for z from our constraint, we have ||h x,y,z i|| = 16 ⇒ p x 2 + y 2 + z 2 = 16 ⇒ z = p 16-x 2-y 2 Therefore, we have the function f ( x,y ) = x + y + p 16-x 2-y 2 to describe the sum. Differentiating: f x ( x,y ) = 1 +-x p 16-x 2-y 2 f y ( x,y ) = 1 +-y p 16-x 2-y 2 If f x ( x,y ) = 0, then 16 = 2 x 2 + y 2 . If f y ( x,y ) = 0, then 16 = x 2 +2 y 2 . Solving the system, we find that the points ( 4 √ 3 , 4 √ 3 ) (-4 √ 3 , 4 √ 3 ) ( 4 √ 3 ,-4 √ 3 ) (-4 √ 3 ,-4 √ 3 ) are critical points for f ....
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