Conept:
f(x, y, z) = x2 + 5y2 + 5z2 - 4x + 40y - 40z + 300
For minimum Value, \(\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = \frac{\partial f}{\partial z} = 0\)
Calculation:
Given:
\(f_x = x^2 - 4x\)
∴ \(\frac{\partial f}{\partial x}\) = 2x - 4 = 0
2x - 4 =0
So, x = 2
Similarly, for \(f_y = 5y^2 + 40y\)
∴ \(\frac{\partial f}{\partial y}\) = 10y + 40 = 0
10y + 40 = 0
So, y = -4
Also, for \(f_z = 5z^2 - 40z\)
∴ \(\frac{\partial f}{\partial z}\) = 10z - 40 = 0
So, z = 4
Now, Putting values of x, y, z in function f(x, y, z),
f(x, y, z) = x2 + 5y2 + 5z2 - 4x + 40y - 40z + 300
f(2, -4, 4) = \(2^2 + 5(-4)^2 + 5(4)^2 - 4(2) + 40(-4) - 40(4) + 300\)
f(2, -4, 4) = 4 + 80 + 80 - 8 - 160 - 160 + 300
f(2, -4, 4) = 136
∴ the minimum value of function f defined by f(x, y, z) = 136
