A space probe in the shape of the ellipsoid 4x^2 + y^2 + 4z^2 = 16 enters the earth's atmosphere and its surface begins to heat. After 1 hour, the temperature at the point (x, y, z... A space probe in the shape of the ellipsoid 4x^2 + y^2 + 4z^2 = 16 enters the earth's atmosphere and its surface begins to heat. After 1 hour, the temperature at the point (x, y, z) on the probe's surface is T(x,y,z) = 8x^2 + 4yz - 16z + 600. Find the hottest point on the probe's surface. Read more

Steps to Solve

1. Define the Ellipsoid Constraint The surface of the ellipsoid is given by the equation: $$ 4x^2 + y^2 + 4z^2 = 16 $$ We can express this in a more manageable form: $$ \frac{x^2}{4} + \frac{y^2}{16} + \frac{z^2}{4} = 1 $$

2. Set up the Temperature Function The temperature function is given by: $$ T(x, y, z) = 8x^2 + 4yz - 16z + 600 $$ Our goal is to maximize this function given the constraint from the ellipsoid.

3. Use the Method of Lagrange Multipliers Introduce a Lagrange multiplier $\lambda$, and set up the equations: $$ \nabla T = \lambda \nabla g $$ where $g(x, y, z) = 4x^2 + y^2 + 4z^2 - 16 = 0$.

4. Compute the Gradients Calculate the gradients:

   • For temperature: $$ \nabla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z} \right) = (16x, 4z, 4y - 16) $$
   • For the constraint: $$ \nabla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z} \right) = (8x, 2y, 8z) $$

5. Set Up the Lagrange Equations Set the Lagrange equations from the gradients: $$ 16x = \lambda(8x) $$ $$ 4z = \lambda(2y) $$ $$ 4y - 16 = \lambda(8z) $$

6. Solve for Variables in Terms of $\lambda$ From the first equation, if $x \neq 0$: $$ \lambda = 2 $$

   Substitute $\lambda = 2$ into the other equations:

   • From the second equation: $$ 4z = 2(2y) \implies z = y $$
   • From the third equation: $$ 4y - 16 = 2(8z) = 16z \implies 4y - 16 = 16y \implies 12y = 16 \implies y = \frac{4}{3} $$

7. Find $z$ and Substitute Back Since $z = y$, we have: $$ z = \frac{4}{3} $$ Substitute $y$ and $z$ into the ellipsoid equation to find $x$: $$ 4x^2 + \left(\frac{4}{3}\right)^2 + 4\left(\frac{4}{3}\right)^2 = 16 $$ This simplifies to: $$ 4x^2 + \frac{16}{9} + \frac{64}{9} = 16 $$ $$ 4x^2 + \frac{80}{9} = 16 \implies 4x^2 = 16 - \frac{80}{9} \implies 4x^2 = \frac{144 - 80}{9} = \frac{64}{9} $$ Thus: $$ x^2 = \frac{16}{9} \implies x = \frac{4}{3} \text{ or } x = -\frac{4}{3} $$

8. Evaluate Both Points The candidates are: $$ (x, y, z) = \left(\frac{4}{3}, \frac{4}{3}, \frac{4}{3}\right) \text{ and } \left(-\frac{4}{3}, \frac{4}{3}, \frac{4}{3}\right) $$ Substitute these points back into the temperature function to find the maximum temperature.

9. Calculate the Temperature For $ \left(\frac{4}{3}, \frac{4}{3}, \frac{4}{3}\right) $: $$ T\left(\frac{4}{3}, \frac{4}{3}, \frac{4}{3}\right) = 8\left(\frac{4}{3}\right)^2 + 4\left(\frac{4}{3}\right)\left(\frac{4}{3}\right) - 16\left(\frac{4}{3}\right) + 600 $$ For $ \left(-\frac{4}{3}, \frac{4}{3}, \frac{4}{3}\right) $, the temperature will be the same due to symmetry.

10. Final Calculation Calculate both expressions to determine the hottest point.