Find the maximum and minimum distance of the point (3, 4, 12) from the sphere z²+x²+y²=1 using Lagrange’s multiplier method.
Answers
Answer:
Minimum distance = 12, Maximum distance = 14.
Step-by-step explanation:
Using Lagrange’s multiplier method to find the points on sphere which are near and far is as follow.
See the attached diagram, point A is near to P and Point B is far from P.
Identify the objective function and the constraint equation.
The objective is to find the maximum/minimum distance between any point on sphere and the point (3,4,12)
The center of sphere x² + y² + z² = 1 is at (0,0,0)
Radius = 1 unit.
d² = (x−3)²+(y−4)²+(z-12)²
So the objective function is
f(x,y,z) = d² = (x−3)²+(y−4)²+(z-12)²
The constraint equation is that the point (x,y,z) must be on the given sphere. Rewriting this as g(x,y,z)=0 gives
g(x,y,z) = x²+y²+z² – 1 = 0
According to the method of Lagrange multipliers, the maximum/minimum values occur when
∇f(x,y,z) =λ∇g(x,y,z) (λ – is the distance from point)
Differentiation of F and G functions, we get
⟨2(x−3), 2(y−4), 2(z-12)⟩ = λ⟨2x, 2y, 2z⟩
This along with the constraint equation gives us the following system of four equations
2(x − 3) = 2λx
2(y − 4) = 2λy
2(z -12) = 2λz
x²+y²+z² =1
x−λx = 3
x(1−λ) = 3
x = 3/(1−λ)
Similarly, we get y = 4/(1−λ)
z = 12/(1−λ)
Substituting these into the last equation gives
x²+y²+z²=1
3²/(1−λ)²+4²/(1−λ)²+12²/(1−λ)²=1
169 = (1 – λ)²
1 – λ = 13, -13
λ = -14, 12
x = 3/(1−λ) = 3/15 or 3/11 = 1/5 or 3/11
y = 4/(1−λ) = 4/15 or 4/11
z = 12/(1−λ) = 12/15 or 12/11 = 4/5 or 12/11
The closest and farthest points are (1/5, 4/15, 4/5) and (3/11, 4/11, 12/11)
Distance between (3,4,12) and above points is 14 and 12.
Hence minimum distance = 12, Maximum distance = 14.
find the maximum and minimum distance of the point (3,4,12) from the sphere z2+X2+y2=1 using Lagrange multiplier method