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..
z = x^2 + y^2, using Lagrange Multipliers Method. Thanks for your help..
-
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.
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.