Respuesta :
Since a $5 decrease in price increases customers by 20 we can say that we have two points:
(125,100) and (120,120), from these we can find the slope or rate of change of customers as a function of price...
m=20/-5
m=-4
m=-4
c(p)=-4p+b, now we can use (125,100) to solve for b
100=-4(125)+b
100=-500+b
600=b, so our number of customers as a function of price is:
c(p)=600-4p
Revenue will simply be the number of customers times the price charged per customer...or p*c(p):
r(p)=600p-4p^2
We can find price that creates maximum revenue by finding when the derivative is equal to zero...
dr/dp=600-8p
dr/dp=0 only when
0=600-8p
8p=600
p=75
So the price that maximizes revenue is $75.
(125,100) and (120,120), from these we can find the slope or rate of change of customers as a function of price...
m=20/-5
m=-4
m=-4
c(p)=-4p+b, now we can use (125,100) to solve for b
100=-4(125)+b
100=-500+b
600=b, so our number of customers as a function of price is:
c(p)=600-4p
Revenue will simply be the number of customers times the price charged per customer...or p*c(p):
r(p)=600p-4p^2
We can find price that creates maximum revenue by finding when the derivative is equal to zero...
dr/dp=600-8p
dr/dp=0 only when
0=600-8p
8p=600
p=75
So the price that maximizes revenue is $75.
The linear relationship is c(p) = -4p + b. Then the price that maximizes revenue is $75.
What is the linear system?
It is a system of an equation in which the highest power of the variable is always 1.
A carpet cleaning business has 100 customers when they charge $125 for cleaning.
Research shows that every $5 reduction in price attracts another 20 customers.
Since a $5 decrease in price increases customers by 20, we can say that we have two options.
(120, 100) and (120, 120) from these we can find the slope or rate of change of customers as a function of price.
[tex]\rm m = \dfrac{20}{-5}\\\\ m = -4[/tex]
Then we have
[tex]\rm c(p ) = -4 p +b[/tex]
Now, we can use (125, 100) to solve for b
100 = -4(125) + b
b = 600
So our number of customers as a function of price is
[tex]\rm c(p) = 600 - 4p[/tex]
Revenue will simply be the number of customers times the price charged per customer will be
[tex]\rm r(p) = 600 p -4p^2[/tex]
We can find a price that creates maximum revenue by finding when the derivative is equal to zero.
[tex]\rm \dfrac{dr}{dp} = 600 - 8p\\\\[/tex]
WE know that for maximum,
[tex]\rm \dfrac{dr}{dp} = 0\\\\0 = 600 - 8p\\\\p = 75[/tex]
So, the price that maximizes revenue is $75.
More about the linear system link is given below.
