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:
$$q(m)=am(am+b)^{\frac{3}{2}}$$
The total revenue from $q$ units is:
$$R(q)=\frac{cq}{(dq+e)^{\frac{1}{2}}}$$
where $$0<a,b,c,d,e$$.
Given that revenue is price per unit time units sold, then price per unit $p$ is:
(1) $$p=\frac{R}{q}=\frac{c}{(dq+e)^{\frac{1}{2}}}$$
Find $$\frac{dR}{dq}$$
(2) $$\frac{dR}{dq}=\frac{(dq+e)^{\frac{1}{2}}(c)-(cq)\left(\frac{1}{2}(dq+e)^{-\frac{1}{2}}(d) \right)}{\left((dq+e)^{\frac{1}{2}} \right)^2}=\frac{c(dq+2e)}{2(dq+e)^{\frac{3}{2}}}$$
Find $$\frac{dR}{dm}$$.
Using the chain rule, we may state:
$$\frac{dR}{dm}=\frac{dR}{dq}\cdot\frac{dq}{dm}$$
$$\frac{dq}{dm}=(am)\left(\frac{3}{2}(am+b)^{\frac{1}{2}}(a) \right)+(a)\left((am+b)^{\frac{3}{2}} \right)=a(am+b)^{\frac{1}{2}}\left(\frac{5}{2}am+b \right)$$
And so we have:
(3) $$\frac{dR}{dm}=\left(\frac{c(dq+2e)}{2(dq+e)^{\frac{3}{2}}} \right)\left(a(am+b)^{\frac{1}{2}}\left(\frac{5}{2}am+b \right) \right)$$
Next, we want to identify the data given in this particular problem:
$$a=1.8,\,b=18,\,c=1312,\,d=5,\,e=24660$$
To answer the questions, we need to find the value of $q$, when $m=10$:
$$q(10)=18(18+18)^{\frac{3}{2}}=18\cdot6^3=3888$$
a) Using formula (1), we find the price per unit $p$ is:
$$p=\frac{1312}{(5(3888)+24660)^{\frac{1}{2}}}=\frac{656}{105}$$
To two decimal places of accuracy, this is:
$$p\approx6.25$$
b) Using formula (2), we find the marginal revenue is:
$$\frac{dR}{dq}=\frac{1312\left(5(3888)+2(24660) \right)}{2\left(5(3888)+24660 \right)^{\frac{3}{2}}}=\frac{125296}{25725}$$
To two decimal places of accuracy, this is:
$$\frac{dR}{dq}\approx4.87$$
c) Using formula (3), we find the marginal-revenue product is:
$$\frac{dR}{dm}=\frac{125296}{25725}\left(\frac{9}{5}\left(\frac{9}{5}(10)+18 \right)^{\frac{1}{2}}\left(\frac{5}{2}\left(\frac{9}{5}(10) \right)+18 \right) \right)=\frac{20297952}{6125}$$
To two decimal places of accuracy, this is:
$$\frac{dR}{dm}\approx3313.95$$
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.
$$\frac{\Delta R}{\Delta m}=\frac{R(q(11))-R(q(10))}{11-10}=R(q(11))-R(q(10))$$
Now, we need to compute:
$$q(11)=\frac{99}{5}\left(\frac{99}{5}+18 \right)^{\frac{3}{2}}=\frac{56133}{25}\sqrt{\frac{21}{5}}$$
And so we find:
$$\frac{\Delta R}{\Delta m}=R\left(\frac{56133}{25}\sqrt{\frac{21}{5}} \right)-R(3888)\approx27651.8824207754-\frac{850176}{35}\approx3361.13956363254$$
Compare this with the approximation we found:
$$\frac{\Delta R}{\Delta m}\approx\frac{20297952}{6125}\approx3313.95134693878$$
We find our estimate from part c) is close to the actual value, in fact there is about a 1.4% difference.
