Find the shortest distance between the plane and paraboloid
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Find the shortest distance between the plane and paraboloid

[From: ] [author: ] [Date: 11-05-27] [Hit: ]
x²+y² = 4x+2y−20 → (x−2)²+(y−1)² = −15 which is a contradiction.The perp distance of (x,y,z) from the plane is | (4x+2y−z−20)/√21 |.L(x,y,......
Find the shortest distance between the plane 4x+2y -z = 20 and the paraboloid
z = x^2 + y^2, using Lagrange Multipliers Method. Thanks for your help..

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Note the plane and paraboloid do not meet, for if they did then

x²+y² = 4x+2y−20 → (x−2)²+(y−1)² = −15 which is a contradiction.


The perp distance of (x,y,z) from the plane is | (4x+2y−z−20)/√21 |.

Choose objective function as (4x+2y−z−20)² … ( /21 can be added later )

The constraint is x²+y²−z = 0

L(x,y,z,λ) = (4x+2y-z -20)² − λ(x²+y²−z)

Optimality conditions :

8(4x+2y-z -20) − 2λx = 0 … (i)
4(4x+2y-z -20) − 2λy = 0 … (ii)
−2(4x+2y-z -20) + λ = 0 … (iii)
x²+y²−z = 0 … (iv)

From (i), (ii) & (iii) : 2λx = 4λy = 4λ … (v)

If λ=0 then from (iii) 4x+2y−z−20=0 and (x,y,z) lies on the plane.
Since the plane and paraboloid do not meet there is no solution with (iv).

Hence λ≠0 and from (v) x=2 & y=1

Using (iv) z = 5 and | (4x+2y−z−20)/√21 | = 15/√21

Because of squared nature of objective function, clearly a minimum.
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