Máximos e mínimos
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Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.7 Maximum and Minimum Valuesf = y 36x 9x 72xy 2 36 9x 36y
3
2
(
2
2 1/2
)
2 x . Setting f =0 gives y=0 or y = x 2
2 2
2
but y>0 , so only the latter solution
4 2 1 2 or x= , y= and then z =(36 12 12)/4=3 . y 3 3 3 The fact that this gives a maximum volume follows from the geometry. This maximum volume is 2 16 1 . V =(2x)(2y)(2z)=8 ( 3 )= 3 3 3 applies. Substituting this y into f =0 gives x =
( a2b2c2 b2c2x2 a2c2y2) 1/2 44. Here maximize f (x,y)=xy ab f =yc x 2 2 2 2 2
. Then a b 2a y b x
2 2 2 2 2 2 2 2 2 1/2
a b 2b x a y
2 2
2 2
2 2 2 2 2 1/2
ab abc bc x ac y
2 2 2 2
(
2 2 2
2 2 2
)
and f =xc y 2 2 2
ab abc bc x ac y y (
2 2 2
2 2 2
)
. Then f =0 x (with x , y>0 ) implies y = y=
a b 2b x a
2 2
and substituting into f =0 implies 3b x =a b or x=
2 2
2 2
1 a, 3
1 1 b and then z= c . Thus the maximum volume of such a rectangle is 3 3 8 V =(2x)(2y)(2z)= abc . 3 3 xy (6 x 2y) , then the maximum volume is V =xyz . 3 1 1 1 2 f = 6y 2xy y = y(6 2x 2y) and f = x ( 6 x 4y ) . Setting f =0 and f =0 gives the critical x 3 y 3 x y 3 point ( 2,1 ) which geometrically must yield a maximum. Thus the volume of the largest such box is 2 4 V =(2)(1) = . 3 3 45. Maximize f (x,y)=
(
)
46. Surface area =2(xy+xz+yz)=64 cm , so xy+xz+yz=32 or z= 32 xy 2 32 2xy x 32y 2xy x y f (x,y)=xy =y . Then f = x 2 2 x+y (x+y) (x+y)
2 2 3 2 2 2
2
32 xy . Maximize the volume x+y
2
and f =x y 32 2xy y (x+y)
2
2
. Setting f =0 x 32 x 2 2 2 2 2 32 x 4x 32 x =0 or implies y= and substituting into f =0 gives 32 4x y 2x 8 64/3 8 8 4 2 2 2 64 3x +64x (32) =0 . Thus x = = or x= , y= and z= . Thus the box is a cube 6 6 16/ 6 6 6
( )(
)( ) (
)
22
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.7 Maximum and Minimum Values
with edge length
8 cm. 6
47. Let