Question

The temperature of a point \((x, y, z)\) on the unit sphere is given by $$ T(x, y, z)=1+x y+y z $$ By using the method of Lagrange multipliers find the temperature of the hottest point on the sphere.

Short answer

The temperature of the hottest point on the sphere is \( \frac{4}{3} \)

Step-by-step solution

- Define the Constraint

The problem involves a unit sphere, so the constraint is given by the equation of the sphere: equation: \( g(x, y, z) = x^2 + y^2 + z^2 - 1 = 0 \)

- Set Up the Lagrange Function

Introduce the Lagrange multiplier \( \lambda \) and set up the Lagrange function \( \mathcal{L}(x, y, z, \lambda) = T(x, y, z) + \lambda g(x, y, z) \)Substitute the given functions:\( \mathcal{L}(x, y, z, \lambda) = 1 + xy + yz + \lambda (x^2 + y^2 + z^2 - 1) \)

- Find the Partial Derivatives

Compute the partial derivatives of \( \mathcal{L} \) with respect to \( x, y, z \) and set them to zero:\( \frac{\partial \mathcal{L}}{\partial x} = y + 2\lambda x = 0 \)\( \frac{\partial \mathcal{L}}{\partial y} = x + z + 2\lambda y = 0 \)\( \frac{\partial \mathcal{L}}{\partial z} = y + 2\lambda z = 0 \)\( \frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 + z^2 - 1 = 0 \)

- Solve the System of Equations

Solve the system of equations derived from the partial derivatives:1. \( y + 2\lambda x = 0 \)2. \( x + z + 2\lambda y = 0 \)3. \( y + 2\lambda z = 0 \)4. \( x^2 + y^2 + z^2 - 1 = 0 \)Using equation (1) and (3), we get: \( y = -2\lambda x \) and \( y = - 2\lambda z \) Thus, \( x = z \) Substitute into (2): \( x + x + 2\lambda (-2\lambda x) = 0 \) This simplifies to: \( 2x (1 - 2\lambda^2) = 0 \) So either \( x = 0 \) or \( \lambda = \pm \frac{1}{\sqrt{2}} \)

- Analyze Critical Points

Substitute the solutions back to find the critical points: If \( x = 0 \) then \( y = 0 \) or \( z = \pm 1 \). This gives points (0,0,1) and (0,0,-1) If \( \lambda = \pm \frac{1}{\sqrt{2}} \), we have x = z and \( y = \mp \frac{x}{\sqrt{2}} \)Use constraint: \( x^2 + (\mp \frac{x}{\sqrt{2}})^2 + x^2 = 1 \) gives \( x^2 + \frac{x^2}{2} + x^2 = 1 \) Thus, \( x = \pm \frac{1}{\sqrt{3}} \)

- Calculate Temperature

Substitute the critical points into the temperature function to find the maximumPoints: (0,0,1), (0,0,-1), (1/√3,-1/√6, 1/√3), (-1/√3, 1/√6, -1/√3)Calculate T(0,0,1) = 1, T(0,0,-1) = 1, T(1/√3, -1/√6, 1/√3) = 1, and for -1/√3 Thus, hottest point - T(x,y,z) = 1 + 1/3 = 4/3

Key concepts

unit sphere constraint

The unit sphere constraint is essential in optimization problems where the solution must lie on a spherical surface. The equation of a unit sphere is given by:
\[ x^2 + y^2 + z^2 = 1 \]
This equation means that any point \( (x, y, z) \) on the unit sphere is at a distance of 1 from the origin.
In the context of Lagrange multipliers, the unit sphere acts as a constraint that restricts the set of possible solutions.
To incorporate this constraint into our optimization problem, we define a constraint function:
\[ g(x, y, z) = x^2 + y^2 + z^2 - 1 = 0 \]
This function represents the difference between the left-hand side of the unit sphere equation and 1. Setting it to zero ensures we remain on the unit sphere.
It is crucial to understand that every solution we find must satisfy this constraint equation to be valid.

temperature function

The temperature function describes how temperature varies at each point \( (x, y, z) \) on the sphere. In this exercise, it is given by:
\[ T(x, y, z) = 1 + xy + yz \]
The goal is to find the maximum temperature on the sphere by optimizing this function under the unit sphere constraint.
When dealing with functions like this, remember each term indicates a relationship between the variables.
For example, \( xy \) indicates that the temperature depends not only on individual coordinates but also on their products.
By finding the critical points of this function, under the given constraint, we determine the points where the temperature could be maximized.

partial derivatives

Partial derivatives measure how a multivariable function changes as one particular variable changes, holding others constant.
For Lagrange multipliers, we take the partial derivatives of the Lagrangian function with respect to each variable and set them to zero.
This helps in finding critical points.
In our problem, the Lagrangian is given by: \[ \mathcal{L}(x, y, z, \lambda) = T(x, y, z) + \lambda(x^2 + y^2 + z^2 - 1) \]
We find the partial derivatives with respect to \( x, y, z, \lambda \):
\[ \frac{\partial \mathcal{L}}{\partial x} = y + 2\lambda x = 0 \]
\[ \frac{\partial \mathcal{L}}{\partial y} = x + z + 2\lambda y = 0 \]
\[ \frac{\partial \mathcal{L}}{\partial z} = y + 2\lambda z = 0 \]
\[ \frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 + z^2 - 1 = 0 \]
Solving these equations simultaneously helps us identify the critical points of the temperature function, respecting the unit sphere constraint.
Understanding partial derivatives is key to applying the Lagrange multiplier method correctly.