S3TAR3H's question at Yahoo Answers regarding marginal revenue

[MarkFL]

Here is the question:

Need help with derivatives please!?


All answers involve a unit of dollars, so you must enter your answers accurate to two decimal places!
A factory owner who employes m workers finds that they produce

q= 1.8m(1.8m+18)^3/2 units of product per day.

The total revenue R in dollars is

R=1312q / (24660+5q)^1/2

(a) From the fact that
revenue =(price per unit)*(number of units)
it follows that
R=(price per unit)*q

So when there are 10 workers, the price per unit is ? dollars.

(b) When there are 10 workers, the marginal revenue is ? dollars/(one unit of product).

(c) The marginal-revenue product is defined as the rate of change of revenue with respect to the number of employees. Therefore,
marginal-revenue product=dR/dm

If q and R are given as above then, when m= 10, the marginal-revenue product is ? dollars/(one worker). This means that if employee number 11 is hired, revenue will increase by approximately ? dollars per day.

I tried substituting 10 for m and my answer was 3888 which is wrong :(
also I know that the marginal revenue is the derivative of the revenue function but i still can't seem to find the right answer :(

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello S3TAR3H,

Let's work this problem using parameters rather than numbers, so that we then have formulas we can use in the future for similar problems, as I have see this same problem type posted many times in my years on math help forums.

The number of units produced by $m$ workers is:

\(\displaystyle q(m)=am(am+b)^{\frac{3}{2}}\)

The total revenue from $q$ units is:

\(\displaystyle R(q)=\frac{cq}{(dq+e)^{\frac{1}{2}}}\)

where \(\displaystyle 0<a,b,c,d,e\).

Given that revenue is price per unit time units sold, then price per unit $p$ is:

(1) \(\displaystyle p=\frac{R}{q}=\frac{c}{(dq+e)^{\frac{1}{2}}}\)

Find \(\displaystyle \frac{dR}{dq}\)

(2) \(\displaystyle \frac{dR}{dq}=\frac{(dq+e)^{\frac{1}{2}}(c)-(cq)\left(\frac{1}{2}(dq+e)^{-\frac{1}{2}}(d) \right)}{\left((dq+e)^{\frac{1}{2}} \right)^2}=\frac{c(dq+2e)}{2(dq+e)^{\frac{3}{2}}}\)

Find \(\displaystyle \frac{dR}{dm}\).

Using the chain rule, we may state:

\(\displaystyle \frac{dR}{dm}=\frac{dR}{dq}\cdot\frac{dq}{dm}\)

\(\displaystyle \frac{dq}{dm}=(am)\left(\frac{3}{2}(am+b)^{\frac{1}{2}}(a) \right)+(a)\left((am+b)^{\frac{3}{2}} \right)=a(am+b)^{\frac{1}{2}}\left(\frac{5}{2}am+b \right)\)

And so we have:

(3) \(\displaystyle \frac{dR}{dm}=\left(\frac{c(dq+2e)}{2(dq+e)^{\frac{3}{2}}} \right)\left(a(am+b)^{\frac{1}{2}}\left(\frac{5}{2}am+b \right) \right)\)

Next, we want to identify the data given in this particular problem:

\(\displaystyle a=1.8,\,b=18,\,c=1312,\,d=5,\,e=24660\)

To answer the questions, we need to find the value of $q$, when $m=10$:

\(\displaystyle q(10)=18(18+18)^{\frac{3}{2}}=18\cdot6^3=3888\)

a) Using formula (1), we find the price per unit $p$ is:

\(\displaystyle p=\frac{1312}{(5(3888)+24660)^{\frac{1}{2}}}=\frac{656}{105}\)

To two decimal places of accuracy, this is:

\(\displaystyle p\approx6.25\)

b) Using formula (2), we find the marginal revenue is:

\(\displaystyle \frac{dR}{dq}=\frac{1312\left(5(3888)+2(24660) \right)}{2\left(5(3888)+24660 \right)^{\frac{3}{2}}}=\frac{125296}{25725}\)

To two decimal places of accuracy, this is:

\(\displaystyle \frac{dR}{dq}\approx4.87\)

c) Using formula (3), we find the marginal-revenue product is:

\(\displaystyle \frac{dR}{dm}=\frac{125296}{25725}\left(\frac{9}{5}\left(\frac{9}{5}(10)+18 \right)^{\frac{1}{2}}\left(\frac{5}{2}\left(\frac{9}{5}(10) \right)+18 \right) \right)=\frac{20297952}{6125}\)

To two decimal places of accuracy, this is:

\(\displaystyle \frac{dR}{dm}\approx3313.95\)

As a check, we can find compute the change in revenue over the change in the number of workers when going from 10 to 11 workers, and compare it with the estimate we just found.

\(\displaystyle \frac{\Delta R}{\Delta m}=\frac{R(q(11))-R(q(10))}{11-10}=R(q(11))-R(q(10))\)

Now, we need to compute:

\(\displaystyle q(11)=\frac{99}{5}\left(\frac{99}{5}+18 \right)^{\frac{3}{2}}=\frac{56133}{25}\sqrt{\frac{21}{5}}\)

And so we find:

\(\displaystyle \frac{\Delta R}{\Delta m}=R\left(\frac{56133}{25}\sqrt{\frac{21}{5}} \right)-R(3888)\approx27651.8824207754-\frac{850176}{35}\approx3361.13956363254\)

Compare this with the approximation we found:

\(\displaystyle \frac{\Delta R}{\Delta m}\approx\frac{20297952}{6125}\approx3313.95134693878\)

We find our estimate from part c) is close to the actual value, in fact there is about a 1.4% difference.