Respuesta :
Answer:
The price that maximizes the revenue is $15 per book.
Step-by-step explanation:
We have a demand that has a linear relation to price.
The two points we will use to calculate the demand function are:
- Price p=10, Demand q=100
- Price p=30, Demand q=0
We have the equation:
[tex]q=m\cdot p+b[/tex]
Then, we replace:
[tex]q(30)=0=m(30)+b\\\\q(10)=100=m(10)+b\\\\\\b=-30m\\\\100=10m+(-30m)=-20m\\\\m=-100/20=-5\\\\b=-30(-5)=150\\\\\\q(p)=-5p+150[/tex]
The revenue R can be calculated as the multiplication of price and quantity.
We can maximize R by deriving and equal that to 0:
[tex]R=p\cdot q=p(-5p+150)=-5p^2+150p\\\\\\\dfrac{dP}{dp}=-5(2p)+150=0\\\\\\-10p+150=0\\\\p=150/10=15[/tex]
The price that maximizes the revenue is $15 per book.
The price that will offer the maximum revenue would be as follows:
$15
Find the price
What information do we have,
When Price = $10, Q = 100 Units
When Price = $ 30, Q = 0 units
With the above information, we can take the equation:
Q = Maximum (p + b)
by putting the values in both the above situations,
30 = 0 = m30 + b
10 = 0 = m10 + b
we get b = -30m. by putting this, we get
100 = 10m + (-30m) = -20m
∵ q(p) = -5p + 150
Through the equation and putting the above values, we get:
[tex]R = p.q.[/tex]
[tex]= p(-5p + 150)[/tex]
[tex]= -5p^2 + 150p[/tex]
$15 as the maximum price.
Thus, $ 15 is the correct answer.
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