There's an error in your integration. I would revisit that.
A mathematician is selling goods at a car boot sale. She believes that the rate at
which she makes sales depends on the length of time since the start of the sale,
t hours, and the total value of sales she has made up to that time, £ x.
She uses the model
where k is a constant.
Given that after two hours she has made sales of £96 in total,
(a) solve the differential equation and show that she made £72 in the first hour of
The mathematician believes that is it not worth staying at the sale once she is making sales at a rate of less than £10 per hour.
(b) Verify that at 3 hours and 5 minutes after the start of the sale, she should have
t= 0, x =0 , c = 0
t = 2 , x = 96 k = 576
I really suck on part'b', how would I go about it?
Plugging in the value of and solving for at the appropriate time was what was required. A simple step that you've already done in your head, no doubt; however, in order to show it, you just need to write it down.
As to the integral, it's a situation with an unclear notation. The correct integration is , not
. See the difference?
As for part b), try finding out what her sales rate is at the given time.