Answer :
Answer:
The minimum distance of √((195-19√33)/8) occurs at ((-1+√33)/4; (-1+√33)/4; (17-√33)/4) and the maximum distance of √((195+19√33)/8) occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)
Step-by-step explanation:
Here, the two constraints are
g (x, y, z) = x + y + 2z − 8
and
h (x, y, z) = x ² + y² − z.
Any critical point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we actually don’t need to find an explicit equation for the ellipse that is their intersection.
Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)
Then the distance from (x, y, z) to the origin is given by
√((x − 0)² + (y − 0)² + (z − 0)² ).
This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema of the square of the distance. Thus, our objective function is
f(x, y, z) = x ² + y ² + z ²
and
∇f = (2x, 2y, 2z )
λ∇g = (λ, λ, 2λ)
µ∇h = (2µx, 2µy, −µ)
Thus the system we need to solve for (x, y, z) is
2x = λ + 2µx (1)
2y = λ + 2µy (2)
2z = 2λ − µ (3)
x + y + 2z = 8 (4)
x ² + y ² − z = 0 (5)
Subtracting (2) from (1) and factoring gives
2 (x − y) = 2µ (x − y)
so µ = 1 whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0 into (3) gives us 2z = −1 and thus z = − 1 /2 . Subtituting z = − 1 /2 into (4) and (5) gives us
x + y − 9 = 0
x ² + y ² + 1 /2 = 0
however, x ² + y ² + 1 /2 = 0 has no solution. Thus we must have x = y.
Since we now know x = y, (4) and (5) become
2x + 2z = 8
2x ² − z = 0
so
z = 4 − x
z = 2x²
Combining these together gives us 2x² = 4 − x , so
2x² + x − 4 = 0 which has solutions
x = (-1+√33)/4
and
x = -(1+√33)/4.
Further substitution yeilds the critical points
((-1+√33)/4; (-1+√33)/4; (17-√33)/4) and
(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).
Substituting these into our objective function gives us
f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8
f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8
Thus minimum distance of √((195-19√33)/8) occurs at ((-1+√33)/4; (-1+√33)/4; (17-√33)/4) and the maximum distance of √((195+19√33)/8) occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)
In this question, we have 2 constraints:
The plane g ( x , y , z ) = x + y + 2 z - 8
The paraboloid h ( x , y , z ) = x² + y² - z
We need to apply Lagrange Multipliers to answer it
The solution are:
The nearest point P = ( -9.06/4 , -9.06 /4 , 6.23 )
The farthest point Q ( (7.06) /4 , (7.06) /4 , 10.26)
The Objective Function (F) is the distance between the ellipse and the origin, In this case, we don´t need to know the equation of the ellipse
The Objective Function is
F = √ x² + y² + z² and as this function has the same critical points that
F = x² + y² + z² we will use this one
Then:
δF/δx = 2×x δF/δy = 2×y δF/δz = 2×z
λ ×δg/δx = λ λ δg/δy = λ λ δg/δz = 2× λ
μ× δh/δx = 2× μ×x μ× δh/δy = 2× μ×y μ× δh/δz= - μ
Therefore we get our five equations.
2×x = λ + 2× μ×x (1)
2×y = λ + 2× μ×y (2)
2×z = 2× λ - μ (3)
x + y + 2 z - 8 = 0 (4)
x² + y² - z = 0 (5)
Subtracting equation (2) from equation (1)
2×x - 2×y = 2× μ×x - 2× μ×y
( x - y ) = μ × ( x - y ) then μ = 1 and by substitution in eq. (2)
2×y = λ + 2×y then λ = 0
From eq. (3)
2×z = - 1 z = -1/2
By subtitution in eq. (4) and (5)
x + y - 1 - 8 = 0 ⇒ x + y = 9
x² + y² + 1/2 = 0 this equation has no solution.
If we make x = y
Equation (4) and (5) become
2× x + 2× z = 8
2×x² - z = 0 ⇒ z = 2×x²
2× x + 4×x² = 8 ⇒ 2×x² + x - 8 = 0
Solving for x x₁,₂ = ( -1 ± √ 1 + 64 ) / 4
x₁,₂ = ( -1 ± √65 ) 4
x₁ = (-1 + √65) /4 x₂ = ( -1 - √65) /4 √ 65 = 8.06
x₁ = 1.765 x₂ = - 2.265
And z = 2×x² ⇒ z₁ = 6.23 z₂ =
And critical points are:
P ( x₁ y₁ z₁ ) ( (7.06) /4 , (7.06) /4 , 6.23 )
Q ( x₂ y₂ z₂ ) ( -9.06/4 , -9.06 /4 , 10.26 )
And by simple inspeccion we see That
minimum distance is the point P = ( -9.06/4 , -9.06 /4 , 6.23 )
the point Q ( (7.06) /4 , (7.06) /4 , 10.26) is the farthest point
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