Question

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

Short answer

The hottest point is (2,2,2) and the temperature there is 8.

Step-by-step solution

- Identify the function and the constraint

The temperature function is given by \[ T(x, y, z) = x y z. \] The constraint is the equation of the sphere: \[ g(x, y, z) = x^2 + y^2 + z^2 - 12 = 0. \]

- Apply the method of Lagrange multipliers

To find the points that maximize or minimize T subject to the constraint, set up the Lagrange function \[ L(x,y,z, \lambda) = x y z + \lambda (x^2 + y^2 + z^2 - 12). \] Take partial derivatives of L and set them equal to zero: \[ \begin{cases} \frac{\partial L}{\partial x} = yz + \lambda \cdot 2x = 0, \ \frac{\partial L}{\partial y} = xz + \lambda \cdot 2y = 0, \ \frac{\partial L}{\partial z} = xy + \lambda \cdot 2z = 0, \ \frac{\partial L}{\partial \lambda} = x^2 + y^2 + z^2 - 12 = 0. \end{cases} \]

- Solve the system of equations

From \( \frac{\partial L}{\partial x} = yz + 2\lambda x = 0 \), you get \( yz = -2 \lambda x \). Similarly, \( xz = -2 \lambda y \) and \( xy = -2 \lambda z \). note that if any variable is zero, the only feasible solution considering the constraint will be \( x=y=z=0 \), which does not satisfy sphere constraint since its \( x^2 + y^2 + z^2 = 0 eq 12.\)

- Consider non-trivial solutions

Assume none of x, y, or z is zero. Divide the equations pairwise: \( \frac{yz}{xz} = \frac{ -2\lambda x}{-2\lambda y} \rightarrow \frac{y}{x} = \frac{x}{y} \rightarrow y^2 = x^2 \rightarrow y = \pm x \). Similarly, divide another set: \( yz / xy = z / y \rightarrow x^2 = z^2 \rightarrow z = \pm x \).

- Find feasible solutions

Options: (x, x, x), (-x, -x, -x), (-x, x, x), etc. Go back to the constraint: \(3x^2 = 12 \rightarrow x^2 = 4 \rightarrow x = \pm 2 \) thus (2,2,2), (-2,-2,-2). Compute temperature T in both \[ T(2,2,-2)=8 \] \[ T(-2,-2,-2)=-8\]

- Choose the hot point

The temperature at points \((2,2,-2)=8\)

Key concepts

Temperature Function

In multivariable calculus, we often work with functions that take multiple variables as input and produce a single output. In our problem, the temperature at any given point \( (x, y, z) \) is represented by the function \ T(x, y, z) = xyz \. This function tells us how the temperature varies in the 3-dimensional space as the values of \( x \), \( y \), and \( z \) change.
To understand this function better, think of it as a rule that assigns a specific temperature to each point based on the coordinates. For example, if \( x = 2 \), \( y = 3 \), and \( z = 4 \), then \ T(2, 3, 4) = 2 \times 3 \times 4 = 24 \.
The temperature function is key for finding the hottest or coldest points in this space, especially when we're restricted by certain conditions, like being on the surface of a sphere.

Constraint Equation

A constraint equation represents a condition that limits the possible solutions to a problem. In our exercise, the constraint is that we must find the temperature only on the surface of the sphere.
This sphere is described by the equation \ x^2 + y^2 + z^2 = 12 \. This equation ensures that any point \( (x, y, z) \) satisfying it lies exactly on the surface of a sphere with a radius related to 12.
When we introduce constraints in optimization problems, we usually can't use regular methods to find maximum or minimum values. Instead, we use techniques like Lagrange multipliers to incorporate the constraints directly into our calculations.

Sphere Equation

The given sphere equation \ x^2 + y^2 + z^2 = 12 \ defines a set of all points \( (x, y, z) \) that are at a fixed distance from the origin. You can visualize this as a 3D ball centered at the origin.
To draw this sphere, every point on its surface satisfies the constraint \ x^2 + y^2 + z^2 = 12 \. For example, the point \( (2, 2, 2) \) lies on the sphere because \ 2^2 + 2^2 + 2^2 = 12 \.
This surface constraint means we consider only those combinations of \( x, y, \) and \( z \) that lie exactly on this sphere, not inside or outside.

System of Equations

To apply Lagrange multipliers, we create a system of equations. This involves forming the Lagrange function \ L(x, y, z, \lambda) = xyz + \lambda (x^2 + y^2 + z^2 - 12) \.
Next, we take partial derivatives of \ L \ with respect to each variable: \ x, y, z \, and \lambda \, and set these derivatives equal to zero.
This step gives us a system of equations:

• \ \frac{\partial L}{\partial x} = yz + \lambda \cdot 2x = 0 \
• \ \frac{\partial L}{\partial y} = xz + \lambda \cdot 2y = 0 \
• \ \frac{\partial L}{\partial z} = xy + \lambda \cdot 2z = 0 \
• \ \frac{\partial L}{\partial \lambda} = x^2 + y^2 + z^2 - 12 = 0 \

Solving this system gives us critical points \( (x, y, z) \) where the temperature can be maximized or minimized, subject to the constraint.

Feasible Solutions

To find the feasible or valid solutions, we need to consider non-trivial cases.
When solving the system of equations, we found that the feasible solutions occur when \ y = \pm x \ and \ z = \pm x \. Using these relations, we substitute back into the sphere equation \ x^2 + y^2 + z^2 = 12 \.
Substituting \ x = 2 \, we get the points \ (2, 2, 2) \ and \ (-2, -2, -2) \.
Calculating the temperature at these points, we find:

• \ T(2, 2, 2) = 2 \cdot 2 \cdot 2 = 8 \
• \ T(-2, -2, -2) = -2 \cdot -2 \cdot -2 = -8 \

Hence, the maximum temperature occurs at \( (2, 2, 2) \) with a temperature of 8 units.