A space probe in the shape of the ellipsoid 4x² + y² + 4z² = 16 enters the earth's atmosphere and its surface begins to heat. After 1 hour the temperature at the point (x, y, z) on... A space probe in the shape of the ellipsoid 4x² + y² + 4z² = 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) = 8x² + 4yz - 16z + 600. Find the hottest point on the probe's surface. Read more

Steps to Solve

1. Define the functions

   The temperature function is given by

   $$ T(x, y, z) = 8x^2 + 4yz - 16z + 600. $$

   The constraint from the ellipsoid is

   $$ g(x, y, z) = 4x^2 + y^2 + 4z^2 - 16 = 0. $$

2. Set up the Lagrange multipliers

   We need to solve the equations derived from the method of Lagrange multipliers, which require that the gradients of the temperature function and the constraint are proportional. This means we need to find:

   $$ \nabla T = \lambda \nabla g $$

   where

   • ( \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) )

3. Calculate the gradients

   Calculate the partial derivatives for ( T ) and ( g ):

   • For ( T ):

     • ( \frac{\partial T}{\partial x} = 16x )
     • ( \frac{\partial T}{\partial y} = 4z )
     • ( \frac{\partial T}{\partial z} = 4y - 16 )

   • For ( g ):

     • ( \frac{\partial g}{\partial x} = 8x )
     • ( \frac{\partial g}{\partial y} = 2y )
     • ( \frac{\partial g}{\partial z} = 8z )

4. Set up the system of equations

   Now we set up the equation from the gradients:

   $$ \begin{align*} 16x &= \lambda (8x) \quad (1)\ 4z &= \lambda (2y) \quad (2)\ 4y - 16 &= \lambda (8z) \quad (3)\ 4x^2 + y^2 + 4z^2 - 16 &= 0 \quad (4) \end{align*} $$

5. Solve the equations

   From equation (1), if ( x \neq 0 ):

   $$ \lambda = 2. $$

   Substitute ( \lambda ) into equations (2) and (3):

   • From (2):

   $$ 4z = 2(2y) \implies z = y. $$

   • From (3):

   $$ 4y - 16 = 16z \implies 4y - 16 = 16y \implies 12y = 16 \implies y = \frac{4}{3}. $$

   Back-substituting into ( z = y ):

   $$ z = \frac{4}{3}. $$

   Substitute ( y ) into ( g(x,y,z) = 0):

   $$ 4x^2 + \left( \frac{4}{3} \right)^2 + 4\left( \frac{4}{3} \right)^2 - 16 = 0. $$

   Computing gives:

   $$ 4x^2 + \frac{16}{9} + \frac{64}{9} - 16 = 0 \implies 4x^2 - \frac{80}{9} = 0 $$

   Which leads to

   $$ 4x^2 = \frac{80}{9} \implies x^2 = \frac{20}{9} \implies x = \pm\frac{2\sqrt{5}}{3}. $$

6. Find the temperature at the points

   Now, we can find the corresponding temperature ( T ) at the points:

   • For ( ( \frac{2\sqrt{5}}{3}, \frac{4}{3}, \frac{4}{3}) ) and ( ( -\frac{2\sqrt{5}}{3}, \frac{4}{3}, \frac{4}{3}) ), substitute into the temperature function:

   $$ T = 8\left(\frac{2\sqrt{5}}{3}\right)^2 + 4\left(\frac{4}{3}\right)\left(\frac{4}{3}\right) - 16\left(\frac{4}{3}\right) + 600. $$

   This evaluates to the same value for both.

7. Evaluate T to find max

   Compute:

   $$ T = 8\left(\frac{20}{9}\right) + 4\left(\frac{16}{9}\right) - \frac{64}{3} + 600. $$

   Simplifying,

   $$ T = \frac{160}{9} + \frac{64}{9} - \frac{192}{9} + 600 = \frac{32}{9} + 600. $$

   Combine terms:

   $$ T = \frac{32}{9} + 5400/9 = \frac{5432}{9}. $$