d= d = but if (x, y, z) lies on the plane x + 2y + z = 9, then z = -x-2y+9 x-1 x-1 X = We can minimize d by minimizing the simpler expression d²= f(x, y) = x-1 3)² + y² + (18- minimum at (x, y) = and y d= + y² + (z+ 9)², By solving the equations fx2(x - 1) 2(18-x-2y) = 4x + 4y - 38 fy2y4(18-x-2y) = 4x+10y - 72 we find that the only critical point is (x, y) = + y² + (18-x-2y)². x-1 y2+(18-x-2y)². = 0, ])² + ( ²7 )² + (¹7) ²³ = 0 and Since fxx = 4, fxy = 4, and fyy=10, we have D(x, y) = fxxyy (fxy)² = 24> 0 and fxx > 0, so by the second derivatives test / has a local 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, -9). If then we get the following. + y² + (18-x-2y)² and so we have The shortest distance from (1, 0, -9) to the plane x + 2y + z = 9 is
d= d = but if (x, y, z) lies on the plane x + 2y + z = 9, then z = -x-2y+9 x-1 x-1 X = We can minimize d by minimizing the simpler expression d²= f(x, y) = x-1 3)² + y² + (18- minimum at (x, y) = and y d= + y² + (z+ 9)², By solving the equations fx2(x - 1) 2(18-x-2y) = 4x + 4y - 38 fy2y4(18-x-2y) = 4x+10y - 72 we find that the only critical point is (x, y) = + y² + (18-x-2y)². x-1 y2+(18-x-2y)². = 0, ])² + ( ²7 )² + (¹7) ²³ = 0 and Since fxx = 4, fxy = 4, and fyy=10, we have D(x, y) = fxxyy (fxy)² = 24> 0 and fxx > 0, so by the second derivatives test / has a local 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, -9). If then we get the following. + y² + (18-x-2y)² and so we have The shortest distance from (1, 0, -9) to the plane x + 2y + z = 9 is
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 12T
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Transcribed Image Text:The distance from any point (x, y, z) to the point (1, 0, -9) is
1
)² + y² + (z + 9)²,
d =
but if (x, y, z) lies on the plane x + 2y + z = 9, then z =
d =
x-1
We can minimize d by minimizing the simpler expression
d² = f(x, y) =
x-1
)² + y² + (18
minimum at (x, y) =
X =
d =
By solving the equations
fx = 2(x-1) 2(18-x-2y) = 4x + 4y - 38
fy= 2y 4(18-x-2y) = 4x+10y - 72
we find that the only critical point is (x, y) =
3² + y² + (
and y =
+ y² + (18-x-2y)².
1
-x-2y +9
+(18-x-2y)².
+ (17) ²
= 0,
= 0 and
. Since fxx = 4, fxy = 4, and fyy = 10, we have D(x, y) = fxxyy (fxy)² = 24> 0 and fxx > 0, so by the second derivatives test f has a local
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, -9). If
then we get the following.
+y²+(18-x-2y)²
and so we have
The shortest distance from (1, 0, -9) to the plane x + 2y + z = 9 is
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