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

Steps to Solve

1. Identify the temperature function

The temperature function is given by ( T(x, y, z) = 8x^2 + 4yz - 16z + 600 ).

2. Understand the ellipsoid equation

The equation of the ellipsoid is ( 4x^2 + y^2 + 4z^2 = 16 ).

To rewrite this in a standard form, we can divide the entire equation by 16: $$ \frac{x^2}{4} + \frac{y^2}{16} + \frac{z^2}{4} = 1. $$

3. Use the method of Lagrange multipliers

Since we're trying to find the maximum temperature on the surface defined by the ellipsoid, we set up the Lagrangement multiplier equation:

• Minimize ( f(x, y, z) = -T(x, y, z) )
• Subject to the constraint ( g(x, y, z) = 4x^2 + y^2 + 4z^2 - 16 = 0 )

4. Set the gradients equal

Compute the gradients: $$ \nabla f = (-16x, 4z, 4y - 16) $$ $$ \nabla g = (8x, 2y, 8z) $$ Set them equal to each other, introducing the Lagrange multiplier ( \lambda ): $$ (-16x, 4z, 4y - 16) = \lambda (8x, 2y, 8z) $$

5. Formulate the system of equations

This gives us a system of equations: $$ -16x = 8\lambda x \tag{1} $$ $$ 4z = 2\lambda y \tag{2} $$ $$ 4y - 16 = 8\lambda z \tag{3} $$ Along with the constraint ( g(x, y, z) = 0 ).

6. Solve the system

From equation (1): If ( x \neq 0 ), then ( -16 = 8\lambda \Rightarrow \lambda = -2 ).

Using ( \lambda = -2 ) in equations (2) and (3): Substituting into equation (2): $$ 4z = -4y \Rightarrow z = -y \tag{4} $$

Substituting equation (4) into equation (3): $$ 4y - 16 = -16z \Rightarrow 4y - 16 = -16(-y) \Rightarrow 4y - 16 = 16y $$ Thus, $$ -12y = 16 \Rightarrow y = -\frac{4}{3} \tag{5} $$

7. Find ( x, z )

Using ( y = -\frac{4}{3} ) in equation (4): $$ z = \frac{4}{3} $$ Substituting ( y ) and ( z ) back to the ellipsoid equation to find ( x ): $$ 4x^2 + \left(-\frac{4}{3}\right)^2 + 4\left(\frac{4}{3}\right)^2 = 16 $$ $$ 4x^2 + \frac{16}{9} + \frac{64}{9} = 16 $$ $$ 4x^2 = 16 - \frac{80}{9} $$ $$ 4x^2 = \frac{144 - 80}{9} = \frac{64}{9} \Rightarrow x^2 = \frac{16}{9} \Rightarrow x = \pm \frac{4}{3} $$

8. Calculate the temperature at these points

Now we have the candidates:

• ( \left( \frac{4}{3}, -\frac{4}{3}, \frac{4}{3} \right) )
• ( \left( -\frac{4}{3}, -\frac{4}{3}, \frac{4}{3} \right) )

Substituting these into the temperature function to find the hottest point.