A container in R³ has the shape of the cube given by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1. A plate is placed in the container such that it occupies that portion of the plane x + y + z =... A container in R³ has the shape of the cube given by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1. A plate is placed in the container such that it occupies that portion of the plane x + y + z = 1 that lies on the cubic container. If the container is heated so that the temperature at each point (x, y, z) is given by T(x, y, z) = 4 - 2x² - y² - z² in hundreds of degrees Celsius, what are the hottest and coldest points on the plate? Use Lagrange's multipliers method to identify all the critical points in the interior of the plate and be certain to examine the plate boundaries. Read more

Steps to Solve

1. Define the constraint and temperature function

The temperature function is given by: $$ T(x, y, z) = 4 - 2x^2 - y^2 - z^2 $$

The constraint function based on the plane is: $$ g(x, y, z) = x + y + z - 1 = 0 $$

2. Set up the Lagrange function

We need to set up the Lagrange function: $$ \mathcal{L}(x, y, z, \lambda) = T(x, y, z) + \lambda g(x, y, z) $$

Substituting the functions: $$ \mathcal{L}(x, y, z, \lambda) = (4 - 2x^2 - y^2 - z^2) + \lambda (x + y + z - 1) $$

3. Find the gradients

We need to find gradients: $$ \nabla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z} \right) $$ $$ \nabla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z} \right) $$

Calculating the partial derivatives:

• For $x$: $\frac{\partial T}{\partial x} = -4x$
• For $y$: $\frac{\partial T}{\partial y} = -2y$
• For $z$: $\frac{\partial T}{\partial z} = -2z$
• For the Lagrange constraint: $\nabla g = (1, 1, 1)$

4. Set the equations based on Lagrange multipliers

From Lagrange's multipliers, we get: $$ \nabla T = \lambda \nabla g $$

This gives us the system of equations:

1. $-4x = \lambda$

2. $-2y = \lambda$

3. $-2z = \lambda$

4. Along with the constraint: $x + y + z = 1$

5. Solve the system of equations

From the equations:

• From equations (1) and (2), we have $-4x = -2y \implies 2y = 4x \implies y = 2x$
• From (1) and (3): $-4x = -2z \implies 2z = 4x \implies z = 2x$

Substituting these into the constraint: $$ x + 2x + 2x = 1 \implies 5x = 1 \implies x = \frac{1}{5} $$

Thus, substituting back:

• $y = \frac{2}{5}$
• $z = \frac{2}{5}$

6. Calculate the temperature at critical point

The critical temperature is obtained using: $$ T\left(\frac{1}{5}, \frac{2}{5}, \frac{2}{5}\right) = 4 - 2\left(\frac{1}{5}\right)^2 - \left(\frac{2}{5}\right)^2 - \left(\frac{2}{5}\right)^2 $$

Calculating: $$ = 4 - 2\left(\frac{1}{25}\right) - \left(\frac{4}{25}\right) - \left(\frac{4}{25}\right) $$ $$ = 4 - \frac{2}{25} - \frac{4}{25} - \frac{4}{25} = 4 - \frac{10}{25} = 4 - \frac{2}{5} = \frac{20}{5} - \frac{2}{5} = \frac{18}{5} $$

7. Evaluate the boundaries of the plate

Evaluate $T(x,y,z)$ on boundaries where one variable is 0. For example:

• When $x = 0$: $$ y + z = 1 \implies T(0, y, z) = 4 - y^2 - z^2 $$
• Use this for boundaries iteratively, substituting the values, and checking minimums and maximums.

8. Check temperatures at corners of the cube

Evaluate $T$ at corners of the cube within the boundaries defined by the plane:

• $(0, 1, 0)$, $(0, 0, 1)$, $(1, 0, 0)$, etc., and check resulting temperatures.