Q20Mathematics I

Question

Use Lagrange's method to find the maximum and minimum distance of the point $(3, 4, 12)$ from the sphere

x^{2} + y^{2} + z^{2} = 1

Answer 1

The Method of Lagrange Multipliers is an incredibly elegant and powerful mathematical strategy used in multivariable calculus for finding the absolute local maxima and minima of a function subject to strict equality constraints. In this geometric problem, we must find the maximum and minimum physical distances from a specific fixed point in 3D space, $P (3, 4, 12)$ , to any point $(x, y, z)$ lying exactly on the surface of a unit sphere.

Step 1: Defining the Objective and Constraint Functions

The fundamental distance between any point $(x, y, z)$ and the fixed point $(3, 4, 12)$ is given by the 3D distance formula $D = (x - 3)^{2} + (y - 4)^{2} + (z - 12)^{2}$ . To completely avoid the algebraic nightmare of differentiating a complex square root, we employ a standard optimization trick: we instead optimize the square of the distance. Because squaring is a strictly monotonically increasing function for positive values, optimizing $D^{2}$ will mathematically yield the exact same coordinate points as optimizing $D$ .

Thus, our strict objective function to be extremized is:

f (x, y, z) = (x - 3)^{2} + (y - 4)^{2} + (z - 12)^{2}

The point $(x, y, z)$ cannot be anywhere in space; it is absolutely constrained to lie on the surface of a unit sphere centered at the origin. Therefore, our strict constraint function is:

g (x, y, z) = x^{2} + y^{2} + z^{2} - 1 = 0

Step 2: Constructing the Lagrange Equations

The Method of Lagrange Multipliers dictates that at the precise points of extremum, the gradient vectors of the objective function and the constraint function must be perfectly parallel to each other. Mathematically, this is expressed as $\nabla f = λ \nabla g$ , where $λ$ (lambda) is an unknown scalar known as the Lagrange multiplier.

We meticulously calculate the gradient of $f$ :

\nabla f = ⟨ 2 (x - 3), 2 (y - 4), 2 (z - 12)⟩

We calculate the gradient of $g$ :

\nabla g = ⟨ 2 x, 2 y, 2 z ⟩

Equating these gradients vector component by component yields a strict system of four equations (including the original constraint):

$2 (x - 3) = λ (2 x) ⟹ x - 3 = λ x$
$2 (y - 4) = λ (2 y) ⟹ y - 4 = λ y$
$2 (z - 12) = λ (2 z) ⟹ z - 12 = λ z$
$x^{2} + y^{2} + z^{2} = 1$

Step 3: Solving the System of Equations

We must solve this non-linear system algebraically. We rearrange the first three equations to explicitly isolate the variables $x, y,$ and $z$ entirely in terms of the unknown multiplier $λ$ :

x - λ x = 3 ⟹ x (1 - λ) = 3 ⟹ x = \frac{3}{1 - λ}

y - λ y = 4 ⟹ y (1 - λ) = 4 ⟹ y = \frac{4}{1 - λ}

z - λ z = 12 ⟹ z (1 - λ) = 12 ⟹ z = \frac{12}{1 - λ}

To find the exact value of $λ$ , we forcefully substitute these three expressions directly into the fundamental constraint equation (Equation 4):

(\frac{3}{1 - λ})^{2} + (\frac{4}{1 - λ})^{2} + (\frac{12}{1 - λ})^{2} = 1

\frac{9 + 16 + 144}{( 1 - λ ) ^{2}} = 1

\frac{169}{( 1 - λ ) ^{2}} = 1 ⟹ (1 - λ)^{2} = 169

Taking the square root of both sides yields two distinct mathematical possibilities for the denominator: $(1 - λ) = 13$ or $(1 - λ) = - 13$ .

Step 4: Finding the Extreme Points and Distances

We now substitute these two possibilities back into our equations for $x, y,$ and $z$ to find the two exact physical points on the sphere where the extrema occur.

Case 1: $(1 - λ) = 13$ This yields the coordinate point $P_{1} = (3/13, 4/13, 12/13)$ . We calculate the absolute distance $D_{1}$ from this point to $(3, 4, 12)$ :

D_{1} = (\frac{3}{13} - 3)^{2} + (\frac{4}{13} - 4)^{2} + (\frac{12}{13} - 12)^{2}

D_{1} = (\frac{- 36}{13})^{2} + (\frac{- 48}{13})^{2} + (\frac{- 144}{13})^{2} = \frac{1}{13} 1296 + 2304 + 20736 = \frac{24336}{13} = \frac{156}{13} = 12

Case 2: $(1 - λ) = - 13$ This yields the coordinate point $P_{2} = (- 3/13, - 4/13, - 12/13)$ . We calculate the absolute distance $D_{2}$ from this point to $(3, 4, 12)$ :

D_{2} = (\frac{- 3}{13} - 3)^{2} + (\frac{- 4}{13} - 4)^{2} + (\frac{- 12}{13} - 12)^{2}

D_{2} = (\frac{- 42}{13})^{2} + (\frac{- 56}{13})^{2} + (\frac{- 168}{13})^{2} = \frac{1}{13} 1764 + 3136 + 28224 = \frac{33124}{13} = \frac{182}{13} = 14

Through rigorous calculus, we have definitively proven that the absolute maximum distance is 14, and the absolute minimum distance is 12.

Q20Mathematics I

Question

Answer

Step 1: Defining the Objective and Constraint Functions

Step 2: Constructing the Lagrange Equations

Step 3: Solving the System of Equations

Step 4: Finding the Extreme Points and Distances