Use Lagrange multipliers to find the shortest distance from the point (5, 0, −7) to the plane x + y + z = 1.

Question

Use Lagrange multipliers to find the shortest distance from the point

(5, 0, −7)

to the plane

x + y + z = 1.

i need help step by step

Answer 1

Minimize here f(x,y,z) equal to (x − 5)² + (y − 0)² + (z − -7)², constrained by

g(x,y,z) equal to x + y + z − 1 = 0.

∇f is [2(x − 5), 2y, 2(z + 7)]; ∇g is [1, 1, 1].

Equate ∇f to λ∇g. That is to say:

2x − 10 = λ;

2y = λ;

2z + 14 = λ.

This stack of equations leads to:

x = (10 + λ)/2;

y = λ/2;

z = (λ − 14)/2.

Place these last 3 expressions into x + y + z = 1 to reach

{10 + λ + λ + λ − 14}/2 = 1 which goes to 3λ − 4 = 2 or λ = 2.

Then take:

x = (10 + 2)/2 or 6;

y = 2/2 or 1;

z = (2 − 14)/2 or -6.

Obtain the distance sought as √[(6 − 5)² + (1 − 0)² + (-6 − -7)²] which reduces to √3.

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As confirmation, minimize the square of the distance between (5, 0, -7) and a chosen

point (x,y,z) that obeys x + y + z = 1.

Set down f(x,y,z) = (x − 5)² + (y − 0)² + (z − -7)².

Consider z a function of x and y and take:

f_x = 2(x − 5) + 2(z + 7)∂z/∂x;

f_y = 2(y − 0) + 2(z + 7)∂z/∂y.

Now differentiate x + y + z = 1 with respect to x & y

to gain:

1 + ∂z/∂x = 0;

1 + ∂z/∂y = 0.

From this last, obtain ∂z/∂x = ∂z/∂y = -1.

Next equate f_x = f_y = 0 to enable:

(x − 5) + (z + 7)(-1) = 0;

(y − 0) + (z + 7)(-1) = 0.

Rewrite as:

x − z = 12;

y − z = 7.

From the last two equations, write

x − y = 5;

z = y − 7.

Finally, gain

x = y + 5;

z = y − 7.

Use x + y + z = 1 and write

(y + 5) + y + (y − 7) or 3y − 2 = 1 which gives y as 1.

This unlocks (x,y,z) as (6,1,-6).

Again, the distance sought is √[(6 − 5)² + (1 − 0)² + (-6 − -7)²] or √3

in agreement with the solution found by Lagrange Multipliers.