A container in R^3 has the shape of the cube given by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1. A plate is placed in the container in such a way that it occupies that portion of the plane x... A container in R^3 has the shape of the cube given by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1. A plate is placed in the container in such a way that it occupies that portion of the plane x + y + z = 1 that lies on the cubical container. If the container is heated so that the temperature at each point (x, y, z) is given by T(x, y, z) = 4 - 2x^2 - y^2 - z^2 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 temperature function and constraint

   The temperature function is given by $$ T(x, y, z) = 4 - 2x^2 - y^2 - z^2 $$ The constraint is the plane defined by the equation $$ g(x, y, z) = x + y + z - 1 = 0 $$

2. Set up Lagrange's multipliers

   We need to find the gradient of the temperature function and the gradient of the constraint:

   • Gradient of $T$: $$ \nabla T = \left( -4x, -2y, -2z \right) $$

   • Gradient of $g$: $$ \nabla g = (1, 1, 1) $$

   According to Lagrange's method, we set $$ \nabla T = \lambda \nabla g $$ This gives us the system of equations: $$ -4x = \lambda $$ $$ -2y = \lambda $$ $$ -2z = \lambda $$

3. Solve for $x$, $y$, and $z$ in terms of $\lambda$

   From the equations, we can express: $$ x = -\frac{\lambda}{4} $$ $$ y = -\frac{\lambda}{2} $$ $$ z = -\frac{\lambda}{2} $$

4. Substitute in the constraint

   Substitute these expressions into the constraint $x + y + z = 1$: $$ -\frac{\lambda}{4} - \frac{\lambda}{2} - \frac{\lambda}{2} = 1 $$ This simplifies to: $$ -\frac{\lambda}{4} - \frac{2\lambda}{4} - \frac{2\lambda}{4} = 1 $$ Adding gives: $$ -\frac{5\lambda}{4} = 1 $$

   Thus, solving for $\lambda$: $$ \lambda = -\frac{4}{5} $$

5. Determine $x$, $y$, and $z$

   Substitute $\lambda$ back to find $x$, $y$, and $z$: $$ x = \frac{1}{5}, \quad y = \frac{2}{5}, \quad z = \frac{2}{5} $$

6. Compute the temperature at critical point

   Now substitute these values into the temperature function: $$ T\left(\frac{1}{5}, \frac{2}{5}, \frac{2}{5}\right) = 4 - 2\left(\frac{1}{5}\right)^2 - \left(\frac{2}{5}\right)^2 - \left(\frac{2}{5}\right)^2 $$

   Calculate step by step: $$ T = 4 - 2\left(\frac{1}{25}\right) - \frac{4}{25} - \frac{4}{25} $$ $$ = 4 - \frac{2}{25} - \frac{4}{25} - \frac{4}{25} $$ $$ = 4 - \frac{10}{25} $$ $$ = 4 - \frac{2}{5} = 4 - 0.4 = 3.6 $$

7. Evaluate temperature on the boundaries of the plate

   The plate boundaries are where one of the variables is at its extreme values (0 or 1). We can check combinations of $(x, y, z)$ that satisfy $x + y + z = 1$:

   • $(1, 0, 0)$
   • $(0, 1, 0)$
   • $(0, 0, 1)$
   • $(1, 0, 0)$ gives $T(1, 0, 0) = 4 - 2(1^2) - 0 - 0 = 2$
   • $(0, 1, 0)$ gives $T(0, 1, 0) = 4 - 0 - 1 - 0 = 3$
   • $(0, 0, 1)$ gives $T(0, 0, 1) = 4 - 0 - 0 - 1 = 3$

8. Compare all temperatures

   Finally, comparing:

   • Critical point: $T\left(\frac{1}{5}, \frac{2}{5}, \frac{2}{5}\right) = 3.6$
   • Boundaries: $T(1, 0, 0) = 2$, $T(0, 1, 0) = 3$, $T(0, 0, 1) = 3$