I'm having trouble with part of a question in a problem set. The question reads as follows:
Continuous price discounting. To encourage buyers to place 100-unit orders, your firm's sales department applies a continuous discount that makes the unit price a function p(x) of the number of units x ordered. The discount decreases the price at the rate of $0.01 per unit ordered. The price per unit for a 100-unit order is p(100)=20.09
a. Find p(x) by solving the following initial value problem:
Differential equation: dp/dx= -1/100p
Initial condition: p(100)= 20.09
My first thought was to take the integral of the differential equation and solve for C like so:
-1/100integral of p= (-1/100)((P^2)/2)= -(P^2)/200 +c
20.09= -(100^2)/200 + C
20.09+50= C
c= 70/09
Therefore p(x)= -(p^2)/200+70.09
But this equation doesn't take the discount into account. So I thought maybe the way to find an equation was to use the equation for continous decay:
P=Ce^kt
Where c= the initial value
k= constant
t= number of units?
So we would have:
P=Ce^-.01t
20.09=Ce^(-.01)(100)
C= 54.61 (rounded)
Therefore P=54.61e^.-01t
But this doesn't incorporate the differential equation at all.
Do either of these equations look correct?
price decreases .01 per unit
dprice/dunit= -.01/unit
dprice=-.01 dunit/unit
price= -.01 ln unit + C
Price(100)=-.01 ln 100+ C
C= 20.09 +.01 ln 100
check that.
To find the correct equation, let's first solve the initial value problem given by the differential equation dp/dx = -1/100 p, with the initial condition p(100) = 20.09.
We can start by separating the variables and integrating both sides of the equation.
dp/p = -1/100 dx
Integrating both sides, we get:
ln|p| = -1/100 x + C
Now, exponentiating both sides, we have:
|p| = e^(-1/100 x + C)
Since we are only interested in the positive value of p, we can drop the absolute value signs:
p = e^(-1/100 x + C)
Since p(100) = 20.09, we can substitute this value into the equation to determine the constant C.
20.09 = e^(-1/100 * 100 + C)
20.09 = e^(-1 + C)
Now we can solve for C. Taking the natural logarithm of both sides, we have:
ln(20.09) = -1 + C
C = ln(20.09) + 1
Substituting this value back into the equation, we get:
p = e^(-1/100 x + ln(20.09) + 1)
Simplifying further, we have:
p = e^(1 - x/100) * 20.09
Therefore, the correct equation that describes the price per unit p as a function of the number of units x ordered is:
p(x) = e^(1 - x/100) * 20.09
This equation takes into account the continuous discount as described in the problem.