At an Oregon fiber-manufacturing facility, an analyst estimates that the weekly number of pounds of acetate fibers that can be produced is given by the function :

Question

At an Oregon fiber-manufacturing facility, an analyst estimates that the weekly number of pounds of acetate fibers that can be produced is given by the function :

z=f(x,y)=1250ln(yx^2)+45(y^2+x)(x^3 -2y)-(xy)^1/2
where
z= the weekly # of pounds of acetate fiber produced
x=the # of skilled workers at the plant
y= the # of unskilled workers at the plant
b) A week later, 20 skilled workers and 12 unskilled workers are employed due to the increasing demand of acetate fiber. What is the rate of change of output with respect to skilled worker?

Answer 1

z = 1250ln(yx^2)+45(y^2+x)(x^3 -2y)-(xy)^1/2

see ∂z/∂x at

http://www.wolframalpha.com/input/?i=1250ln%28yx^2%29%2B45%28y^2%2Bx%29%28x^3+-2y%29-%28xy%29^1%2F2

But you will need some values for x and y if you want a numeric value.

Answer 2

To find the rate of change of output with respect to skilled workers, we need to calculate the partial derivative of the function f(x, y) with respect to x.

The function we're given is:
z = f(x, y) = 1250ln(yx^2) + 45(y^2 + x)(x^3 - 2y) - (xy)^(1/2)

To find the partial derivative of f(x, y) with respect to x, we keep y constant and differentiate with respect to x:

∂z/∂x = (∂/∂x) [1250ln(yx^2) + 45(y^2 + x)(x^3 - 2y) - (xy)^(1/2)]

Let's break it down:

1. The derivative of 1250ln(yx^2) with respect to x is:
1250 * (1/x^2) * (yx^2) = 1250y

2. To differentiate the second term 45(y^2 + x)(x^3 - 2y), we can use the product rule:
(a) The derivative of 45(y^2 + x) with respect to x is simply 45.
(b) The derivative of (x^3 - 2y) with respect to x is 3x^2.

So, the derivative of the second term is: 45 * 3x^2 = 135x^2.

3. The derivative of (xy)^(1/2) with respect to x can be found using the chain rule:
(a) Let u = xy. Then, du/dx = y, and du/dy = x.
(b) Applying the chain rule: (d/dx) [(xy)^(1/2)] = (du/dx) * (d/du) [u^(1/2)]
= y * (1/2) * u^(-1/2)
= (y/2) * (xy)^(-1/2)
= (y/2) * (x^(-1/2))(y^(-1/2))
= (1/2)(xy)^(-1/2)

Now, let's put all the parts together and calculate ∂z/∂x:

∂z/∂x = 1250y + 135x^2 - (1/2)(xy)^(-1/2)

To find the rate of change of output with respect to skilled workers, we substitute the given values of x and y for a week later:

x = 20
y = 12

Now we can calculate the rate of change of output with respect to skilled workers (∂z/∂x):

∂z/∂x = 1250(12) + 135(20^2) - (1/2)(12)(20)^(-1/2)
= 15000 + 54000 - 6√20
≈ 69089.62 pounds per week

Therefore, the rate of change of output with respect to skilled workers is approximately 69089.62 pounds per week.