Question

Calc word problem? A volume discount on a certain item is expressed as a differential equation in which the derivative of the price per item with respect to the number purchased is proportional to the difference between the price and some base price, below which the price can never go. If the price for just one item is $60, and the base price is $30, and the price per item when buying 10 items is $54, what is the price per item when buying 100 items?

Answer 1

If I've set this up right, \[\frac{ dP }{dx }=k(P-P _{0})\]
k is just a constant of proportionality, since dp/dx is proportional to the difference between the price per item P and the base price P0. Now it looks like we can use separation of variables...
\[\frac{ dP }{ P-P _{0} }=k dx\]
\[\frac{ 1 }{ P-P _{0} }dP=k dx\]
integrate both sides
\[\ln (P-P _{0})=kx+C\]
Now we can solve for k and C since they gave us P(1) = $60, P(10) = $54, and the base price P0 = $30.

Answer 2

From the initial conditions we get a system of two equations to solve for k and C:
\[\ln (60-30)=1k+C\]
\[\ln (54-30)=10k+C\]

Answer 3

I'll let you solve the system of equations.

We want to find P(100) ie. the price per item when buying 100 items, so x=100:
\[\ln (P-30)=100k+C\]

Put both sides to the power of e:
\[P-30=e ^{100k+C}\]
\[P=e ^{100k+C}+30\]
and now just insert your values for k and C from above. If I've done everything right, you should get P = $32.58