A firm uses 3 factors of production.Its production function is f(x, y, z)  minx³/y, y², (z⁴  x⁴) /y².If the amount of each input is multiplied by 6, its output will be multiplied by

A) 216.
B) 36.
C) 6.
D) 0.16.
E) The answer depends on the original choice of x, y, and z.

According to the production function, the output (S) is a function of the inputs x, y, and z. If all of the inputs are multiplied by 6, the new production function would be f(6x, 6y, 6z) = S'.
Substituting 6x, 6y, and 6z into the original production function, we get:
S' = 10 min [ (6x / y)^(1/3) * y^(1/3) * (6z)^(1/3) * (1/3) - (6x)^(1/3) * (y / 6z)^(1/3) * (6z)^(1/3) * (1/3) ]
S' = 10 min [ (6^1/3 * x^1/3 * y^1/3 * z^1/3) - (6^1/3 * x^1/3 * y^1/3 * z^1/3) ]
S' = 0
Therefore, if all inputs are multiplied by 6, the output will be zero. This means that the production function exhibits constant returns to scale (i.e., if inputs are multiplied by a certain factor, output is multiplied by the same factor).
To find the factor by which output is multiplied, we can set the inputs to be equal to some constant k (e.g. k=1) and determine the output, and then set the inputs to be equal to 6k and determine the new output. By dividing the new output by the original output, we can determine the factor by which output is multiplied.
Setting x=y=z=k, we get:
S = 10 min [(k/k)^(1/3) * k^(1/3) * k^(1/3) * (1/3) - k^(1/3) * (k/k)^(1/3) * k^(1/3) * (1/3)]
S = 10 min [(1/3)k^(1/3) - (1/3)k^(1/3)]
S = 0
Setting x=y=z=6k, we get:
S' = 10 min [((6k)/(6k))^(1/3) * (6k)^(1/3) * (6k)^(1/3) * (1/3) - (6k)^(1/3) * ((k)/(6k))^(1/3) * (6k)^(1/3) * (1/3)]
S' = 10 min [(1/3)(6k)^(1/3) - (1/3)(k)^(1/3)]
S' = 10 min [(1/3)(6k)^(1/3) - (1/3)(6k)^(1/3)]
S' = 0
So output is also 0 when inputs are multiplied by 6k. Therefore, the production function exhibits constant returns to scale for any choice of x, y, and z, and the factor by which output is multiplied is equal to the factor by which inputs are multiplied, which is 6. Therefore, the answer is (B) 36.