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Thomas Calculus 13th [Solutions]

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1034 Chapter 14 Partial Derivatives<br />

2 2<br />

2x 2y 12x 8y 52 Dx<br />

( x, y) 4x 12 0 and Dy<br />

( x, y) 4y 8 0 critical point is ( 3, 2)<br />

z<br />

13;<br />

2<br />

Dxx ( 3, 2) 4, Dyy ( 3, 2) 4, Dxy ( 3, 2) 0 DxxDyy D xy 16 0 and D xx 0 local<br />

minimum of d( 3, 2, 13) 26<br />

57. V ( x, y, z) (2 x)(2 y)(2 z) 8 xyz;<br />

and<br />

2 2 2 2 2 2 2<br />

x y z 4 z 4 x y V ( x, y) 8xy 4 x y , x 0<br />

2 3<br />

32 y 16x y 8y<br />

32x 16xy 8x<br />

y 0 Vx<br />

( x, y ) 0 and V<br />

2 2<br />

y ( x, y ) 0 critical points are (0, 0),<br />

2 2<br />

4 x y<br />

4 x y<br />

2 , 2 , 2 , 2 , 2 , 2 , and 2 , 2 . Only (0, 0) and 2 , 2 satisfy x 0 and y 0<br />

3 3 3 3 3 3<br />

3 3<br />

3 3<br />

V (0, 0) 0 and V 2 , 2 64 ; On<br />

3 3 3 3<br />

V (0, 0) 0, V (0, 2) 0; On<br />

V (0, 0) 0, V (0, 2) 0; On<br />

2 3<br />

2 2<br />

x 0, 0 y 2 V (0, y) 8(0) y 4 0 y 0, no critical points,<br />

2 2<br />

y 0, 0 x 2 V ( x, 0) 8 x(0) 4 x 0 0, no critical points,<br />

2<br />

2 2 2 2 2<br />

y 4 x , 0 x 2 V x, 4 x 8x 4 x 4 x 4 x 0<br />

no critical points, V (0, 2) 0, V (2, 0) 0. Thus, there is a maximum volume of 64<br />

3 3<br />

2 2 2<br />

3 3 3 .<br />

if the box is<br />

58. S( x, y, z) 2xy 2yz 2 xz;<br />

xyz 27 z 27 S( x, y, z) 2xy 2y 27 2x 27 2 xy 54 54 ,<br />

xy xy xy x y<br />

x 0, y 0; S 54<br />

x ( x , y ) 2 y 0 and S 54<br />

2<br />

y ( x , y ) 2 x<br />

2<br />

x<br />

y<br />

0 Critical point is (3, 3) z 3;<br />

Sxx (3, 3) 4, S yy (3, 3) 4, Dxy (3, 3) 2 DxxDyy 2<br />

D xy 12 0 and D xx 0 local minimum of<br />

S(3, 3, 3) 54<br />

59. Let x height of the box, y width, and z length,<br />

cut out squares of length x from corner of the material<br />

See diagram at right. Fold along the dashed lines to<br />

form the box. From the diagram we see that the length<br />

of the material is 2x y and the width is 2 x z . Thus<br />

2 6 2x<br />

xy<br />

(2 x y)(2 x z) 12 z . Since<br />

2x y<br />

( , , ) ( , )<br />

2xy 6 2x xy<br />

V x y z x y z V x y<br />

2x y<br />

, where<br />

x<br />

0, y 0.<br />

2<br />

2 3 2 2 3<br />

2<br />

2 4 3 2 2<br />

4 3y 4x y 4x y xy<br />

2 12x 4x 4x y x y<br />

Vx<br />

( x, y ) 0 and V ( , ) 0<br />

2<br />

y x y critical points<br />

2<br />

(2 x y)<br />

(2 x y)<br />

are 3, 0 , 3, 0 , 1 , 4 and 1 , 4 . Only 3, 0 and 1 , 4 satisfy x 0 and y 0.<br />

3 3<br />

3 3<br />

3 3<br />

For 3, 0 : z 0;<br />

2<br />

Vxx 3, 0 0, Vyy 3, 0 2 3, Vxy 3, 0 4 3 VxxVyy Vxy<br />

48 0<br />

saddle point. For 1 , 4 : z 4 ; V 1 4 80 1 4 2 1 4 4<br />

3 3 3<br />

xx , , V<br />

3 3 3 3<br />

yy , , V ,<br />

3 3 3 3<br />

xy<br />

3 3 3 3<br />

2<br />

V 16<br />

xxVyy V xy 0 and V 0<br />

3<br />

xx local maximum of V 1 , 4 , 4 16<br />

3 3 3 3 3<br />

Copyright<br />

2014 Pearson Education, Inc.

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