calc 3, Webassign, Thanks Ay Notes O Ask Your Teacher EXAMPLE 5 Find the shortes

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calc 3, Webassign, Thanks

Ay Notes O Ask Your Teacher EXAMPLE 5 Find the shortest distance from the point (1, 0, -5) to the plane x + 2y +z = 16. SOLUTION The distance from any point (x, y, z) to the point (1, 0, -5) Is +y + (z + 5)2 but if (x, y, z) lies on the plane X+ 2y + z = 16, then z = and so we have + y2 + (21 - x - 2y)?. We can minimize d by minimizing the simpler expression d = ffx, y) = ( D +/+ (21 - x - 2y)?. By solving the equations fy = 2(x - 1) - 2(21 - x - 2y) - 4x + 4y - fy = 2y - 4(21 - x - 2y) = 4x + 10y - = 0 we find that the only critical point is (x, y) =( Since fx = 4, fxy = 4, and fyy = 10, we have D(x, y) = fxyy - (Toy)2 = 24 >0 and fxx > 0, so by the Second Derivatives Test f has a local minimum at (x, y) = Intuitively, we can see that this local minimum is actually an absolute minimum because there must be a point on the given plane that is closest to (1, 0, -5). If x= and y = - then + y2 + (21 - x - 2y)? VD?+ (29)² + (19) The shortest distance from (1, 0, -5) to the plane x + 2y + z = 16 is

Explanation / Answer

1 st blank is ( x - 1 )

2 nd blank is z = 16 - x - 2y

3 rd blank is ( x - 1 )

4 th blank is ( x - 1 )

( x -1 )^2 + y^2 + ( 21 - x - 2y)^2

fx = 2 ( x - 1 ) + 0 + 2 ( 21 - x - 2y ) d/dx ( 2 1 - x - 2y)

   = 2x - 2 + ( 42 - 2x - 4y ) ( -1)

   = 2x - 2 - 42 + 2x + 4y

fx = 4x + 4y - 44 = 0

5 th blank is 44

fy = 0 + 2y + 2 ( 21 - x - 2y ) d/dy ( 21 - x - 2y)

   = 2y + ( 42 - 2x - 4y ) ( - 2)

= 2y - 84 + 4x + 8y

fy = 4 x + 10 y - 84 = 0

6 th blank is 84

solve the following

4x + 4y - 44 = 0

4 x + 10 y - 84 = 0   ( subtract )

--------------------------------

0 - 6y + 40 = 0

- 6y = - 40

6y = 40

y = 40 / 6

y = 20 / 3

sub y in

4x + 4y - 44 = 0

4 x + 4 ( 20/3) - 44 = 0

4x = 44 - 80/3

4x = ( 132 - 80 ) / 3

4 x = ( 52 ) / 3

x = ( 52) /12

x = 13 / 3

Critical point is ( x,y) = ( 13/3 , 20/3) --------------> 7 th blank

fxx = 4 , fyy = 10 , fxy = 4

D( 13/3 , 20/3) = fxx( 13/3 , 20/3) * fyy( 13/3 , 20/3) - [ fxy( 13/3 , 20/3) ]^2

                      = 4 * 10 - 4^2

                      = 40 - 16

                     = 24

D( 13/3 , 20/3) > 0 and fxx( 13/3 , 20/3) > 0

f has local minimum at ( x,y) = ( 13/3 , 20/3) -----------> 8 th blank

x = 13 /3   and y = 20/3   --------> 9th and 10 th blanks

11 th blank is ( x - 1 )

x - 1 => ( 13 /3 - 1 )

       => ( 13 - 3 ) / 3

x -1 = 10/3

12 the blank ------->( 10 /3 - 1 ) or 11/3

13 th blank --------> (223/3) or 8.62168

14 th blank --------> 8.62168

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