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minimize (x – w)² + (y – z)² 1,y,w,z subject to ar + by = c+ 1, aw + bz = c – 1.

minimize (x – w)² + (y – z)² 1,y,w,z subject to ar + by = c+ 1, aw + bz = c – 1.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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In this week's notebook, we worked with the equation ax + by = c. (1) As long as at least one of a and b is nonzero, the set of points (x, y) satisfying this equation forms a line in the plane. Next, we considered some parallel lines with equations ax + by = c+1 and ax + by = c – 1. We stated without proof that the distance between these two lines is Va2 + 62 Problem 1. Verify this formula. Here are some steps to get you started. Suppose that (x, y) and (w, z) are arbitrary points on these two lines respectively. In other words, ax + by = c+1 and aw + bz = c – 1. To compute the distance between the lines, we should make these two points as close together as possible. So, we're interested in the constrained opimization problem. minimize (x – w)² + (y z)2 I,y,w,z subject to ax + by = c+1, aw + bz = c – 1. Using our usual trick, we can make things a little simpler by squaring the objective function. That leads to the problem minimize (x – w)² + (y – z)² I,y, w, z subject to ax + by = c+ 1, aw + bz = c –1. Now, use Lagrange multipliers to find the minimum. (Don't forget that you're finding the minimum square distance, so you'll need to take a square root of the minimum value you find to get the minimum distance.)
Transcribed Image Text:In this week's notebook, we worked with the equation ax + by = c. (1) As long as at least one of a and b is nonzero, the set of points (x, y) satisfying this equation forms a line in the plane. Next, we considered some parallel lines with equations ax + by = c+1 and ax + by = c – 1. We stated without proof that the distance between these two lines is Va2 + 62 Problem 1. Verify this formula. Here are some steps to get you started. Suppose that (x, y) and (w, z) are arbitrary points on these two lines respectively. In other words, ax + by = c+1 and aw + bz = c – 1. To compute the distance between the lines, we should make these two points as close together as possible. So, we're interested in the constrained opimization problem. minimize (x – w)² + (y z)2 I,y,w,z subject to ax + by = c+1, aw + bz = c – 1. Using our usual trick, we can make things a little simpler by squaring the objective function. That leads to the problem minimize (x – w)² + (y – z)² I,y, w, z subject to ax + by = c+ 1, aw + bz = c –1. Now, use Lagrange multipliers to find the minimum. (Don't forget that you're finding the minimum square distance, so you'll need to take a square root of the minimum value you find to get the minimum distance.)
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