# Calc 3 Word Problem

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.

[quote]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.

[/quote]

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.

That is some math for your ass right there, nice stuff.

Out source the work to Indian and let them figure it that shit out.

… and here I was thinking powerlifters were stupid

My interpretation:

Now they are using: f(30, 36)=1030(36^(1/2))=1800

Then we add 1 hour to x and calculate y to give us the same result.

1031(y^(1/2))=1800
(y^(1/2))=180/31 |^2
y=33.7148803…
y=33 hours 43 minutes

And change is y-y’=33,71488-36=2 hours 17 minutes.

A: they should decrease the hours of unskilled labor by 2 hours 17 minutes.

[quote]Oroborus wrote:
… and here I was thinking powerlifters were stupid[/quote]

I’ll assume you’re joking. Not like there are members from every professional discipline here, or anything…

[quote]ssplit wrote:
My interpretation:

Now they are using: f(30, 36)=1030(36^(1/2))=1800

Then we add 1 hour to x and calculate y to give us the same result.

1031(y^(1/2))=1800
(y^(1/2))=180/31 |^2
y=33.7148803…
y=33 hours 43 minutes

And change is y-y’=33,71488-36=2 hours 17 minutes.

A: they should decrease the hours of unskilled labor by 2 hours 17 minutes.[/quote]

Doing the math in my head this is the estimated answer that I got.

To remain at the same output while increasing skilled labor the unskilled labors hours must decrease.

[quote]ssplit wrote:
My interpretation:

Now they are using: f(30, 36)=1030(36^(1/2))=1800

Then we add 1 hour to x and calculate y to give us the same result.

1031(y^(1/2))=1800
(y^(1/2))=180/31 |^2
y=33.7148803…
y=33 hours 43 minutes

And change is y-y’=33,71488-36=2 hours 17 minutes.

A: they should decrease the hours of unskilled labor by 2 hours 17 minutes.[/quote]

I’m not quite sure what you’re doing, since you’re wrong and doing it algebraically should look like:

(10)(30)(y^(1/2))=1860

You need to actually find out how much output adding 1 hour of skilled labor costs…

Which gives us y=(1860/300)^2=38.44 hours, or change is 2 hours 26.4 minutes, not what you got.

In any case, the problem asked him to solve this using calculus, which may seem like overkill for such a simple problem, but the point obviously is so he can demonstrate his understanding of calculus, not his ability to solve a simple algebra problem.

[quote]Nanan wrote:
ssplit wrote:
My interpretation:

Now they are using: f(30, 36)=1030(36^(1/2))=1800

Then we add 1 hour to x and calculate y to give us the same result.

1031(y^(1/2))=1800
(y^(1/2))=180/31 |^2
y=33.7148803…
y=33 hours 43 minutes

And change is y-y’=33,71488-36=2 hours 17 minutes.

A: they should decrease the hours of unskilled labor by 2 hours 17 minutes.

Doing the math in my head this is the estimated answer that I got.

To remain at the same output while increasing skilled labor the unskilled labors hours must decrease.

[/quote]

Even if you two are correct and I am wrong about exactly what the problem is asking (who knows, the problem is very poorly worded), this algebra solution isn’t a calculus solution which is what the problem asks for. The calculus solution still goes like I described it, you just have to change a number here or there.

To ssplit:

Sorry, I didn’t realize at first that you were reading the problem differently than I was. Poorly worded problem. Either way we need calculus solution.

Yes haha, I see your point. To be honest I’ve yet to study mathematics in English, so I didn’t know what a calculus solution meant. (I thought I did.) The initial values can still be debated.

This is hardly matters though, we can only hope his test questions are worded better. He should be able to follow your solution anyway.

[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.

[quote]Bujo wrote:

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.[/quote]

Yeah, you could take the problem this way too. I didn’t think of it when I typed out my solution. The past tense of the first sentence hints more towards my understanding, while the present tense of the last sentence hints more towards this understanding. My method will carry over just as well to this way though, you just have to swap out some numbers.

Here’s my take on the problem. I’m guessing this is a linearization problem.

dZ = 10(y)^(1/2) dx + (5x)y^(-1/2) dy.

Given:
dz = 0
x = 30
y = 36
dx = (31-30) = 1
dy = ?

Plug in values and solve for dy = -2.4 hours

[quote]stokedporcupine8 wrote:
Nanan wrote:
ssplit wrote:
My interpretation:

Now they are using: f(30, 36)=1030(36^(1/2))=1800

Then we add 1 hour to x and calculate y to give us the same result.

1031(y^(1/2))=1800
(y^(1/2))=180/31 |^2
y=33.7148803…
y=33 hours 43 minutes

And change is y-y’=33,71488-36=2 hours 17 minutes.

A: they should decrease the hours of unskilled labor by 2 hours 17 minutes.

Doing the math in my head this is the estimated answer that I got.

To remain at the same output while increasing skilled labor the unskilled labors hours must decrease.

Even if you two are correct and I am wrong about exactly what the problem is asking (who knows, the problem is very poorly worded), this algebra solution isn’t a calculus solution which is what the problem asks for. The calculus solution still goes like I described it, you just have to change a number here or there. [/quote]

I was only estimating in my head at about 2hr 20 mins which puts 2:17 and 2:26 well within an acceptable margin of error.

The problem is worded poorly thus confusion is abound… The problem can be solved with alg so I am not totally sure why they ask for a calc3 derived solution.

[quote]Nanan wrote:

The problem is worded poorly thus confusion is abound… The problem can be solved with alg so I am not totally sure why they ask for a calc3 derived solution.
[/quote]

Um, because the point is to test whether the student knows and can apply concepts from calculus? finding calculus problems that test basic concepts without involving a huge amount of work is difficult. They could have asked something like “Find a function that describes the distribution of heat through a 3-D substance as a function of time with the given boundary conditions”, or something like that which would probably involve all sorts of transforms and other fun stuff, but that would be overkill just to see if a student understood the idea of partial differentiation and what it meant.

[quote]stokedporcupine8 wrote:
Nanan wrote:

The problem is worded poorly thus confusion is abound… The problem can be solved with alg so I am not totally sure why they ask for a calc3 derived solution.

Um, because the point is to test whether the student knows and can apply concepts from calculus? finding calculus problems that test basic concepts without involving a huge amount of work is difficult. They could have asked something like “Find a function that describes the distribution of heat through a 3-D substance as a function of time with the given boundary conditions”, or something like that which would probably involve all sorts of transforms and other fun stuff, but that would be overkill just to see if a student understood the idea of partial differentiation and what it meant. [/quote]

I get the chills just thinking about calculus. I am so not glad I am in that class.

[quote]stokedporcupine8 wrote:
Nanan wrote:

The problem is worded poorly thus confusion is abound… The problem can be solved with alg so I am not totally sure why they ask for a calc3 derived solution.

Um, because the point is to test whether the student knows and can apply concepts from calculus? finding calculus problems that test basic concepts without involving a huge amount of work is difficult. They could have asked something like “Find a function that describes the distribution of heat through a 3-D substance as a function of time with the given boundary conditions”, or something like that which would probably involve all sorts of transforms and other fun stuff, but that would be overkill just to see if a student understood the idea of partial differentiation and what it meant. [/quote]

Heh that was a homework problem for one of my physics classes last semester…