Answer :
Answer: a) Q= 5units
b) R= $75
Step-by-step explanation:
The maximum revenue is at dR/dq = 0
Revenue = price x quantity of demand = p × q
Substituting p = 30- 3q
R = (30-3q) × q
R= 30q- 3q2 (q2 = q raised to the power of 2)
dR/dq = 30- 6q = 0
6q = 30
q= 30/6= 5
q= 5 units
R = pq= 30q- 3q2
R= 30(5) - 3( 5×5)
R= 150- 75
R= $75
Goodluck...
Answer:
A) Quantity, q = 5 units.
B) Revenue, r = $75.
Step-by-step explanation:
The demand function for a product is
[tex]p=30-3q[/tex]
where p is the price in dollars when q units are demanded.
Revenue = Price × Quantity
Using the given information, the revenue function is
[tex]R=p\times q[/tex]
[tex]R=(30-3q)\times q[/tex]
[tex]R=30q-3q^2[/tex] .... (1)
Differentiate with respect to q.
[tex]\frac{dR}{dq}=30(1)-3(2q)[/tex]
[tex]\frac{dR}{dq}=30-6q[/tex]
Equate [tex]\frac{dR}{dq}=0[/tex] to find the critical value of q.
[tex]30-6q=0[/tex]
[tex]30=6q[/tex]
Divide both sides by q.
[tex]5=q[/tex]
The value of q is 5.
Differentiate [tex]\frac{dR}{dq}[/tex] with respect to q.
[tex]\frac{d^2R}{dq^2}=-6>0[/tex]
So, the revenue function is maximum at q=5.
Substitute q=5 in equation (1).
[tex]R=30(5)-3(5)^2[/tex]
[tex]R=150-75[/tex]
[tex]R=75[/tex]
Therefore, the level of production that maximizes the total revenue is 5 units and the revenue is $75.