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
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Calculating the Shortest Distance from a Point to a Plane
#### The distance from any point (x, y, z) to the point (1, 0, -9) is
\[ d = \sqrt{(x - 1)^2 + y^2 + (z + 9)^2} \]
#### but if \((x, y, z)\) lies on the plane \( x + 2y + z = 9 \), then \( z = -x - 2y + 9 \) and so we have
\[ d = \sqrt{(x - 1)^2 + y^2 + (18 - x - 2y)^2} \]
#### We can minimize \( d \) by minimizing the simpler expression
\[ d^2 = f(x, y) = (x - 1)^2 + y^2 + (18 - x - 2y)^2 \]
#### By solving the equations
\[ f_x = 2(x - 1) - 2(18 - x - 2y) = 4x + 4y - 38 = 0 \]
\[ f_y = 2y - 4(18 - x - 2y) = 4x + 10y - 72 = 0 \]
#### We find that the only critical point is \((x, y) = \left( \frac{17}{3}, \frac{17}{6} \right) \).
Since \( f_{xx} = 4 \), \( f_{xy} = 4 \), and \( f_{yy} = 10 \), we have
\[ D(x, y) = f_{xx} f_{yy} - (f_{xy})^2 = 24 > 0 \]
and
\[ f_{xx} > 0 \]
So by the second derivatives test \( f \) has a local minimum at \((x, y) = \left( \frac{17}{3}, \frac{17}{6} \right) \).
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 \( x = \frac{17}{3} \) and \( y = \frac{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa2eee30e-7760-4146-894b-912e2799cdb0%2F8fdef301-8083-4e98-a3ad-dad6a6acc906%2Fbem8ng7_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Shortest Distance from a Point to a Plane
#### The distance from any point (x, y, z) to the point (1, 0, -9) is
\[ d = \sqrt{(x - 1)^2 + y^2 + (z + 9)^2} \]
#### but if \((x, y, z)\) lies on the plane \( x + 2y + z = 9 \), then \( z = -x - 2y + 9 \) and so we have
\[ d = \sqrt{(x - 1)^2 + y^2 + (18 - x - 2y)^2} \]
#### We can minimize \( d \) by minimizing the simpler expression
\[ d^2 = f(x, y) = (x - 1)^2 + y^2 + (18 - x - 2y)^2 \]
#### By solving the equations
\[ f_x = 2(x - 1) - 2(18 - x - 2y) = 4x + 4y - 38 = 0 \]
\[ f_y = 2y - 4(18 - x - 2y) = 4x + 10y - 72 = 0 \]
#### We find that the only critical point is \((x, y) = \left( \frac{17}{3}, \frac{17}{6} \right) \).
Since \( f_{xx} = 4 \), \( f_{xy} = 4 \), and \( f_{yy} = 10 \), we have
\[ D(x, y) = f_{xx} f_{yy} - (f_{xy})^2 = 24 > 0 \]
and
\[ f_{xx} > 0 \]
So by the second derivatives test \( f \) has a local minimum at \((x, y) = \left( \frac{17}{3}, \frac{17}{6} \right) \).
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 \( x = \frac{17}{3} \) and \( y = \frac{
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