[quote]stokedporcupine8 wrote:

erik56 wrote:

Can anyone help me with this. I have a test tomorrow and I understand everything except for this one.

Using x hours of skilled labor and y hours of unskilled labor, a manufacturer can produce f(x,y)=10x(y^(1/2)) units. Currently, they have used 30 hours of skilled labor and 36 hours of unskilled and is planning to use 1 additional hour of skilled labor. Use calculus to estimate the corresponding change that they should make in the level of unskilled so that the total output will remain the same.

Exactly what the question is isn’t clear, it would be helpful if you copied the question word for word from the book. (If this is word for word from the book, I feel for you…)

Anyway, I think what you are being asked to do is to "use calculus to estimate the corresponding change that they would have to make in the level of unskilled so that the total output would equal the output of adding 1 hour of skilled labor. Unfortunately, if this is what they’re asking, it’d still be easier to do this with simple algebra, but here’s how (I’d) do it with calculus.

So, assuming they’re saying that given

f(31,36)=1860 units

what y would you need in order to get

f(30,y)=1860

What we need to do is find the change in y that will yield this increase (this requires integrating and partial differentiation.)

Well, in order to answer this you need to know just how many units you gain by adding 1 hour of skilled labor while holding unskilled constant. This is just the partial derivative of f with respect to x, or:

(d/dx)f(x,y)=10(y^(1/2))

Now, since y=36 in the original setup, adding 1 hour of skilled labor when y=36 adds 60 units.

Now for the more complicated bit.

Now we need rate of change for unskilled:

(d/dy)f(x,y)=(5x/y^(1/2)).

Since we’re assuming that we hold skilled constant at 30, change in unskilled, y, is given by 150/y^(1/2).

Now we need to find out what value of y will yield an increase of 60, given this rate of change. This is the integral

Integral from 36 to z of: (150/y^(1/2))dy

With some easy integration, we see that this integral equal 300(sqrt(z) - 6). So now we just set this to 60, the change we got from adding 1 hour of skilled labor and solve for z–which will give us the total number of hours by unskilled labor needed to equal f(31,36) when x=30.

So, given 300(sqrt(z) - 6)=60 and some algebra, we get that z=38.44.

In other words, if we start with 30 hours skilled, 36 hours unskilled, then to get the same effect of adding 1 hour of skilled we need to add 2.44 hours of unskilled.

Anyway, I hope that actually was the question, and if that was the question, I hope what I did was clear and not more complicated than it need have been. It is late though, so I might have not seen an easier way to do this. BTW, some simply algebra checks that this is the correct answer. [/quote]

I agree that this is a poorly worded problem. I agree with your math, but don’t agree with your solution.

I believe the initial set up should be:

f(x,y)=10x(y^(1/2))

Current Factory Operation: Skilled = x = 30 ; Unskilled = y = 36

f(30,36)=1800 units

Proposed Change to Factory Operation: Skilled = x = 31 ; Unskilled = y = ???

f(31,y)=1800 units

This should call for a decrease in unskilled labor.