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EXAMPLE 5 Find the shortest distance from the point (1, 0, -8) to the plane x + 2y + z = 1. SOLUTION The distance from any point (x, y, z) to the point (1, 0, -8) is d = + y? + (z + 8)2 but if (x, y, z) lies on the plane x + 2y + z 1, then z = and so we have d = + y? + (9 – x - 2y)2. We can minimize d by minimizing the simpler express d2 = f(x, y) = )* +v² + (9 - x - 2y)?.

EXAMPLE 5 Find the shortest distance from the point (1, 0, -8) to the plane x + 2y + z = 1. SOLUTION The distance from any point (x, y, z) to the point (1, 0, -8) is d = + y? + (z + 8)2 but if (x, y, z) lies on the plane x + 2y + z 1, then z = and so we have d = + y? + (9 – x - 2y)2. We can minimize d by minimizing the simpler express d2 = f(x, y) = )* +v² + (9 - x - 2y)?.

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter1: Vectors
Section1.3: Lines And Planes
Problem 25EQ
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EXAMPLE 5 Find the shortest distance from the point (1, 0, -8) to the plane x + 2y + z = 1. SOLUTION The distance from any point (x, y, z) to the point (1, 0, -8) is ) + y? + (z + 8)2 d = but if (x, y, z) lies on the plane x + 2y + z = 1, then z = and so we have d = + y2 + (9 – x - 2y)2. We can minimize d by minimizing the simpler expression d? = f(x, y) = + y2 + (9 – x - 2y)². By solving the equations fx = 2(x – 1) - 2(9 – x – 2y) = 4x + 4y – fy = 2y – 4(9 - x - 2y) = 4x + 10y – = 0 = 0 we find that the only critical point is (xX, y) Since fxx = 4, fxy = 4, and fyy = 10, we have D(x, y). fxxfyy - (fxy)? = 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, -8). If X = and y = then )+ y? + (9 - x - 2y)? d ]) · ()* - (4)° 8. 3 The shortest distance from (1, 0, –8) to the plane x + 2y + z = 1 is ||
Transcribed Image Text:EXAMPLE 5 Find the shortest distance from the point (1, 0, -8) to the plane x + 2y + z = 1. SOLUTION The distance from any point (x, y, z) to the point (1, 0, -8) is ) + y? + (z + 8)2 d = but if (x, y, z) lies on the plane x + 2y + z = 1, then z = and so we have d = + y2 + (9 – x - 2y)2. We can minimize d by minimizing the simpler expression d? = f(x, y) = + y2 + (9 – x - 2y)². By solving the equations fx = 2(x – 1) - 2(9 – x – 2y) = 4x + 4y – fy = 2y – 4(9 - x - 2y) = 4x + 10y – = 0 = 0 we find that the only critical point is (xX, y) Since fxx = 4, fxy = 4, and fyy = 10, we have D(x, y). fxxfyy - (fxy)? = 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, -8). If X = and y = then )+ y? + (9 - x - 2y)? d ]) · ()* - (4)° 8. 3 The shortest distance from (1, 0, –8) to the plane x + 2y + z = 1 is ||
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