# Chapter 13 - INSTRUCTORS SOLUTIONS MANUAL SECTION 13.1...

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INSTRUCTOR’S SOLUTIONS MANUAL SECTION 13.1 (PAGE 714) CHAPTER 13. APPLICATIONS OF PARTIAL DERIVATIVES Section 13.1 Extreme Values (page 714) 1. f ( x , y ) = x 2 + 2 y 2 4 x + 4 y f 1 ( x , y ) = 2 x 4 = 0i f x = 2 f 2 ( x , y ) = 4 y + 4 = f y =− 1 . Critical point is ( 2 , 1 ) . Since f ( x , y ) →∞ as x 2 + y 2 , f has a local (and absolute) minimum value at that critical point. 2. f ( x , y ) = xy x + y , f 1 = y 1 , f 2 = x + 1 A = f 11 = 0 , B = f 12 = 1 , C = f 22 = 0. Critical point ( 1 , 1 ) is a saddle point since B 2 AC > 0. 3. f ( x , y ) = x 3 + y 3 3 f 1 ( x , y ) = 3 ( x 2 y ), f 2 ( x , y ) = 3 ( y 2 x ). For critical points: x 2 = y and y 2 = x . Thus x 4 x = 0, that is, x ( x 1 )( x 2 + x + 1 ) = 0. Thus x = 0o r x = 1. The critical points are ( 0 , 0 ) and ( 1 , 1 ) .W eh a v e A = f 11 ( x , y ) = 6 x , B = f 12 ( x , y ) 3 , C = f 22 ( x , y ) = 6 y . At ( 0 , 0 ) : A = C = 0, B 3. Thus AC < B 2 , and ( 0 , 0 ) is a saddle point of f . At ( 1 , 1 ) : A = C = 6, B 3, so AC > B 2 . Thus f has a local minimum value at ( 1 , 1 ) . 4. f ( x , y ) = x 4 + y 4 4 , f 1 = 4 ( x 3 y ), f 2 = 4 ( y 3 x ) A = f 11 = 12 x 2 , B = f 12 4 , C = f 22 = 12 y 2 . For critical points: x 3 = y and y 3 = x . Thus x 9 = x ,o r x ( x 8 1 ) = 0, and x = 0, 1, or 1. The critical points are ( 0 , 0 ) , ( 1 , 1 ) and ( 1 , 1 ) . At ( 0 , 0 ) , B 2 AC = 16 0 > 0, so ( 0 , 0 ) is a saddle point. At ( 1 , 1 ) and ( 1 , 1 ) , B 2 AC = 16 144 < 0, A > 0, so f has local minima at these points. 5. f ( x , y ) = x y + 8 x y f 1 ( x , y ) = 1 y 8 x 2 = f 8 y = x 2 f 2 ( x , y ) x y 2 1 = f x y 2 . For critical points: 8 y = x 2 = y 4 ,so y = r y = 2. f ( x , y ) is not deﬁned when y = 0, so the only critical point is ( 4 , 2 ) .A t ( 4 , 2 ) we have A = f 11 = 16 x 3 1 4 , B = f 12 1 y 2 1 4 , C = f 22 = 2 x y 3 1 . Thus B 2 AC = 1 16 1 4 < 0, and ( 4 , 2 ) is a local maximum. 6. f ( x , y ) = cos ( x + y ), f 1 sin ( x + y ) = f 2 . All points on the lines x + y = n π ( n is an integer) are critical points. If n is even, f = 1 at such points; if n is odd, f 1 there. Since 1 f ( x , y ) 1 at all points in 2 , f must have local and absolute maximum values at points x + y = n π with n even, and local and absolute minimum values at such points with n odd. 7. f ( x , y ) = x sin y . For critical points we have f 1 = sin y = 0 , f 2 = x cos y = 0 . Since sin y and cos y cannot vanish at the same point, the only critical points correspond to x = 0 and sin y = 0. They are ( 0 , n π) , for all integers n . All are saddle points. 8. f ( x , y ) = cos x + cos y , f 1 sin x , f 2 sin y A = f 11 cos x , B = f 12 = 0 , C = f 22 cos y . The critical points are points ( m π, n , where m and n are integers. Here B 2 AC cos ( m cos ( n = ( 1 ) m + n + 1 which is negative if m + n is even, and positive if m + n is odd. If m + n is odd then f has a saddle point at ( m n . If m + n is even and m is odd then f has a local (and absolute) minimum value, 2, at ( m n .I f m + n is even and m is even then f has a local (and absolute) maximum value, 2, at ( m n . 9. f ( x , y ) = x 2 ye ( x 2 + y 2 ) f 1 ( x , y ) = 2 ( 1 x 2 ) e ( x 2 + y 2 ) f 2 ( x , y ) = x 2 ( 1 2 y 2 ) e ( x 2 + y 2 ) A = f 11 ( x , y ) = 2 y ( 1 5 x 2 + 2 x 4 ) e ( x 2 + y 2 ) B = f 12 ( x , y ) = 2 x ( 1 x 2 )( 1 2 y 2 ) e ( x 2 + y 2 ) C = f 22 ( x , y ) = 2 x 2 y ( 2 y 2 3 ) e ( x 2 + y 2 ) .

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## This note was uploaded on 12/16/2009 for the course FEW, FEWEB 400567 taught by Professor Moerdersen during the Fall '09 term at Vrije Universiteit Amsterdam.

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Chapter 13 - INSTRUCTORS SOLUTIONS MANUAL SECTION 13.1...

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