Respuesta :
Answer:
$2.40
Explanation:
Here is the complete question: A deli sells 320 sandwiches per day at a price of $4 each.
A market survey shows that for every $ 0.10 reduction in the price, 40 more sandwiches will be sold. How much should the deli charge in order to maximize revenue?
Given: Number of sandwiches sold per day is 320.
Price of each sandwich is $4.
For every $0.10 reduction in the price, 40 more sandwiches will be sold.
Let´s assume price of each sandwich sold at new reduced price be "x" times.
∴ New price of sandwich is [tex](4-0.10 \times x)[/tex]
As given, for every $ 0.10 reduction in the price, 40 more sandwiches will be sold.
∴ Total number of sandwiches sold at new price= [tex](320+40x)[/tex]
We know, revenue= [tex]price\ of\ sandwiches\times number\ of\ sandwiches[/tex]
⇒ revenue= [tex](4-0.10x)\times (320+40x)[/tex]
using distributive property of multiplication.
⇒ revenue= [tex]-4x^{2} +128x+1280[/tex]
Considering the revenue to be R(x)
[tex]R(x)= -4x^{2} +128x+1280[/tex]
Now taking derivatives of the revenue.
[tex]R(x)[/tex]= [tex]\int\limits {-4x^{2} +128x+1280} \, dx[/tex]
Solving it.
We get, [tex]R(x)= -8x+128[/tex]
Finding critical points by setting derivative equal to 0
R´(x)⇒ [tex]-8x+128= 0[/tex]
Subtracting both side by 128
⇒ [tex]-8x= -128[/tex]
Dividing both side by -8
⇒ [tex]x= \frac{128}{8}[/tex]
∴ x= 16
Hence, price of each sandwich is reduced by $0.10 is 16 times.
Next, finding cost of each sandwich to maximize revenue.
cost of each sandwich to maximize revenue= [tex](4-0.10 \times 16)[/tex]
∴ Cost of each sandwich to maximize revenue= [tex]\$ 2.40[/tex]
Hence, Deli should charges $2.40 for each sandwich to maximize revenue.
