Question

The minimum value of the function \( f(x, y, z) = x^2 + 5y^2 + 5z^2 - 4x + 40y - 40z + 300 \) is _________ (in integer).

Hint:

To find the minimum value of a function, take the partial derivatives, set them equal to zero, and substitute the values back into the original function.

Answer 1

Correct Answer: 136

To find the minimum value of \( f(x, y, z) \), we take the partial derivatives with respect to \( x \), \( y \), and \( z \) and set them equal to zero. The partial derivatives are: \[ \frac{\partial f}{\partial x} = 2x - 4, \quad \frac{\partial f}{\partial y} = 10y + 40, \quad \frac{\partial f}{\partial z} = 10z - 40 \] Setting these equal to zero: \[ 2x - 4 = 0 \implies x = 2 \] \[ 10y + 40 = 0 \implies y = -4 \] \[ 10z - 40 = 0 \implies z = 4 \] Now, substitute \( x = 2 \), \( y = -4 \), and \( z = 4 \) into the original function \( f(x, y, z) \): \[ f(2, -4, 4) = (2)^2 + 5(-4)^2 + 5(4)^2 - 4(2) + 40(-4) - 40(4) + 300 \] \[ = 4 + 5(16) + 5(16) - 8 - 160 - 160 + 300 \] \[ = 4 + 80 + 80 - 8 - 160 - 160 + 300 = 136 \] Thus, the minimum value of the function is \( \boxed{136} \).