15.8%20Prel%20Q%20and%20Ex%201%20-%206

15.8%20Prel%20Q%20and%20Ex%201%20-%206 - 764 C H A P T E R...

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764 CHAPTER 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) We thus showed that if the Fermat point exists, then ^ A < 120 . Similarly, one shows also that ^ B and ^ C must be smaller than 120 . We conclude that if one of the angles in 1 ABC is equal or greater than 120 , then the Fermat point does not exist. In that case, the minimum value of f ( x , y ) occurs at a point where f x or f y do not exist, that is, at one of the points A , B ,or C . 15.8 Lagrange Multipliers: Optimizing with a Constraint (ET Section 14.8) Preliminary Questions 1. Suppose that the maximum of f ( x , y ) subject to the constraint g ( x , y ) = 0 occurs at a point P = ( a , b ) such that f P 6= 0. Which of the following are true? (a) f P is tangent to g ( x , y ) = 0at P . (b) f P is orthogonal to g ( x , y ) = P . SOLUTION (a) Since the maximum of f subject to the constraint occurs at P , it follows by Theorem 1 that f P and g P are parallel vectors. The gradient g P is orthogonal to g ( x , y ) = P , hence f P is also orthogonal to this curve at P . We conclude that statement (b) is false (yet the statement can be true if f P = ( 0 , 0 ) ). (b) This statement is true by the reasoning given in the previous part. 2. Figure 8 shows a constraint g ( x , y ) = 0 and the level curves of a function f . In each case, determine whether f has a local minimum, local maximum, or neither at the labeled point. 4 3 2 1 1 2 3 4 AB g ( x , y ) = 0 g ( x , y ) = 0 f f FIGURE 8 The level curve f ( x , y ) = 2 is tangent to the constraint curve at the point A . A close level curve that intersects the constraint curve is f ( x , y ) = 1, hence we may assume that f has a local maximum 2 under the constraint at A . The level curve f ( x , y ) = 3 is tangent to the constraint curve. However, in approaching B under the constraint, from one side f is increasing and from the other side f is decreasing. Therefore, f ( B ) is neither local minimum nor local maximum of f under the constraint. 3. On the contour map in Figure 9: (a) Identify the points where f = λ g for some scalar . (b) Identify the minimum and maximum values of f ( x , y ) subject to g ( x , y ) = 0. 2 6 2 26 Graph of g ( x , y ) = 0 Contour plot of f ( x , y ) (contour interval 2) 2 4 6 y x FIGURE 9 Contour map of f ( x , y ) ; contour interval 2. (a) The gradient g is orthogonal to the constraint curve g ( x , y ) = 0, and f is orthogonal to the level curves of f . These two vectors are parallel at the points where the level curve of f is tangent to the constraint curve. These are the points A , B , C , D , E in the ±gure:
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SECTION 15.8 Lagrange Multipliers: Optimizing with a Constraint (ET Section 14.8) 765 2 6 2 26 g ( x , y ) = 0 f A , g A A E C D B 2 6 6 (b) The minimum and maximum occur where the level curve of f is tangent to the constraint curve. The level curves tangent to the constraint curve are f ( A ) =− 4 , f ( C ) = 2 , f ( B ) = 6 , f ( D ) 4 , f ( E ) = 4 Therefore the global minimum of f under the constraint is 4 and the global maximum is 6.
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This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

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15.8%20Prel%20Q%20and%20Ex%201%20-%206 - 764 C H A P T E R...

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