Question

The demand function for a product is p=30−3q where p is the price in dollars when q units are demanded. Find the level of production that maximizes the total revenue and determine the revenue. A) Q= units
B) R= $

Answer 1

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 2

Answer:

A) Quantity, q = 5 units.

B) Revenue, r = $75.

Step-by-step explanation:

The demand function for a product is

$p=30-3q$

where p is the price in dollars when q units are demanded.

Revenue = Price × Quantity

Using the given information, the revenue function is

$R=p\times q$

$R=(30-3q)\times q$

$R=30q-3q^2$          .... (1)

Differentiate with respect to q.

$\frac{dR}{dq}=30(1)-3(2q)$

$\frac{dR}{dq}=30-6q$

Equate $\frac{dR}{dq}=0$ to find the critical value of q.

$30-6q=0$

$30=6q$

Divide both sides by q.

$5=q$

The value of q is 5.

Differentiate $\frac{dR}{dq}$ with respect to q.

$\frac{d^2R}{dq^2}=-6>0$

So, the revenue function is maximum at q=5.

Substitute q=5 in equation (1).

$R=30(5)-3(5)^2$

$R=150-75$

$R=75$

Therefore, the level of production that maximizes the total revenue is 5 units and the revenue is $75.