Question

If the temperature at the point \((x, y, z)\) is \(T=x y z\), find the hortest point (or points) on the surface of the sphere \(x^{2}+y^{2}+z^{2}=12\), and find the temperature there.

Short answer

The temperature is 8 or -8.

Step-by-step solution

Identify the objective function and constraint

The objective function to optimize is the temperature function given by \( T = xyz \). The constraint is the surface of the sphere defined by \( x^2 + y^2 + z^2 = 12 \).

Set up the Lagrange function

To solve the problem, introduce a Lagrange multiplier \( \lambda \). The Lagrange function is: \( L(x, y, z, \lambda) = xyz + \lambda (12 - x^2 - y^2 - z^2) \).

Find the partial derivatives and solve the system

Compute the partial derivatives and set them to zero: \( \frac{\partial L}{\partial x} = yz - 2\lambda x = 0 \) \( \frac{\partial L}{\partial y} = xz - 2\lambda y = 0 \) \( \frac{\partial L}{\partial z} = xy - 2\lambda z = 0 \) \( \frac{\partial L}{\partial \lambda} = 12 - x^2 - y^2 - z^2 = 0 \).

Solve the system of equations

From the partial derivatives, {\( yz - 2\lambda x = 0 \) (1), \( xz - 2\lambda y = 0 \) (2), \( xy - 2\lambda z = 0 \) (3), \( 12 - x^2 - y^2 - z^2 = 0 \) (4). From (1), dividing both sides by \( yz \) (assuming \( y eq 0 \) and \( z eq 0 \)), we get \( \lambda = \frac{yz}{2x} \). Using the same approach for (2) and dividing by \( xz \), \( \lambda = \frac{xz}{2y} \). Similarly for (3), \( \lambda = \frac{xy}{2z} \). Equating these: \( \frac{yz}{2x} = \frac{xz}{2y} = \frac{xy}{2z} \).

Simplify and solve

From the above lhs, \( yz/x = xz/y \) implies \( y^2 = x^2 \) or \( y = \pm x \). Similarly, \( x = \pm z \) and \( z = \pm y \). Substituting into the constraint: \( x^2 + y^2 + z^2 = 12 \). \( x^2 + x^2 + x^2 \). Then \( 3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2 \Rightarrow y = \pm 2 \Rightarrow z = \pm 2 \).

Determine the specific points

The points are \((2, 2, 2) ,(-2, -2, -2), (-2, 2, -2), (-2, -2, 2), (2, 2, -2)\).

Calculate temperature at each point

Evaluate \( T = xyz \) at each point: \( (2 \times 2 \times 2) = 8 \) and \((=(-2) \times (-2) \times(-2)= -8).\)

Key concepts

Objective Function

In optimization problems, the objective function is the mathematical expression that you aim to maximize or minimize.
For this exercise, the objective function is the temperature function, given by:
\[ T = xyz \]
Here, your goal is to either maximize or minimize the temperature at any point \(x, y, z\) on a given surface.

Constraint

In this context, a constraint is a condition that the solution must satisfy.
For this problem, the constraint is the surface of a sphere, defined by the equation:
\[ x^2 + y^2 + z^2 = 12 \]
This equation describes a sphere with a radius \( \sqrt{12} \).
Therefore, any point \((x, y, z)\) must lie on this sphere for the solution to be valid.

Partial Derivatives

Partial derivatives measure the rate of change of a multivariable function with respect to one variable, while keeping others constant.
To use Lagrange multipliers, calculate the partial derivatives of the Lagrange function:
\[ L(x, y, z, \lambda) = xyz + \lambda (12 - x^2 - y^2 - z^2) \]
Compute the partial derivatives:
\( \frac{ \partial L}{ \partial x} = yz - 2 \lambda x = 0 \)
\( \frac{ \partial L}{ \partial y} = xz - 2 \lambda y = 0 \)
\( \frac{ \partial L}{ \partial z} = xy - 2 \lambda z = 0 \)
\( \frac{ \partial L}{ \partial \lambda} = 12 - x^2 - y^2 - z^2 = 0 \)
Setting these partial derivatives to zero provides the critical points needed to solve the problem.

System of Equations

A system of equations is a set of equations with multiple variables that need to be solved simultaneously.
Here, you have four equations derived from the partial derivatives:
\( yz - 2 \lambda x = 0 \) (1)
\( xz - 2 \lambda y = 0 \) (2)
\( xy - 2 \lambda z = 0 \) (3)
\( 12 - x^2 - y^2 - z^2 = 0 \) (4)
To solve this system, you can equate the expressions for \lambda from each partial derivative:
\( \frac{ yz}{2x} = \frac{ xz}{2y} = \frac{ xy}{2z} \)
This ultimately leads to relationships like \( y = \pm x \) and \( z = \pm x \).
These relationships, along with equation (4), allow solving for the exact points on the sphere.