Question

The demand equation for a product is p = 45 − 0.01q. Write the revenue as a function of q and find the quantity that maximizes revenue. What price corresponds to this quantity? What is the total revenue at this price?

Answer 1

Answer:

Total revenue at the given price is 50625.

Explanation:

Price p = 45 - 0.01q

Revenue = Demand × Price

$\mathrm{R}(\mathrm{q})=\mathrm{q} \times(45-0.01 \mathrm{q})=45 \mathrm{q}-0.01 \mathrm{q}^{2}$

First derivative of R(q)

$\frac{d R}{d q}=\frac{d}{d q}\left(45 q-0.01 q^{2}\right)$

$=\frac{d}{d q}(45 q)-\frac{d}{d q}\left(0.01 q^{2}\right)$

= 45 - 0.02q

$\text { Set } \frac{d R}{d q}=0$

45 - 0.02q = 0

-0.02q = -45

$\mathrm{q}=\frac{45}{0.02}$

q = 2250

Second derivative of R(q)

$\frac{d^{2} R}{d q^{2}}=\frac{d}{d q}\left(\frac{d R}{d q}\right)$

$=\frac{d}{d q}(45-0.02 q)$

= 0 - 0.02

= - 0.02

Since the second derivative is -ve $\left(\frac{d^{2} R}{d q^{2}}<0\right)$ it can be said that the revenue function has a maximum at q = 2250

The quantity that maximizes the revenue = 2250 items

Price of each item = 45 - 0.01 × 2250  

= 45 - 22.50  

Price of each item is 22.5

$\text { Total revenue }=45(2250)-0.01(2250)^{2}$

= 101250 - 50625

= 50625

Total revenue is 50625.