Answer:
A) Quantity, q = 5 units.
B) Revenue, r = $75.
Step-by-step explanation:
The demand function for a product is
![p=30-3q](https://tex.z-dn.net/?f=p%3D30-3q)
where p is the price in dollars when q units are demanded.
Revenue = Price × Quantity
Using the given information, the revenue function is
![R=p\times q](https://tex.z-dn.net/?f=R%3Dp%5Ctimes%20q)
![R=(30-3q)\times q](https://tex.z-dn.net/?f=R%3D%2830-3q%29%5Ctimes%20q)
.... (1)
Differentiate with respect to q.
![\frac{dR}{dq}=30(1)-3(2q)](https://tex.z-dn.net/?f=%5Cfrac%7BdR%7D%7Bdq%7D%3D30%281%29-3%282q%29)
![\frac{dR}{dq}=30-6q](https://tex.z-dn.net/?f=%5Cfrac%7BdR%7D%7Bdq%7D%3D30-6q)
Equate
to find the critical value of q.
![30-6q=0](https://tex.z-dn.net/?f=30-6q%3D0)
![30=6q](https://tex.z-dn.net/?f=30%3D6q)
Divide both sides by q.
![5=q](https://tex.z-dn.net/?f=5%3Dq)
The value of q is 5.
Differentiate
with respect to q.
![\frac{d^2R}{dq^2}=-6>0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2R%7D%7Bdq%5E2%7D%3D-6%3E0)
So, the revenue function is maximum at q=5.
Substitute q=5 in equation (1).
Therefore, the level of production that maximizes the total revenue is 5 units and the revenue is $75.