1. ## profit, percentage, differentiation

The manager of a supermarket usually adds a mark-up of 20% to the wholesale prices of all the goods he sells. He reckons that he has a loyal core of F customers and that, if he lowers his mark-up to x% he will attract an extra k(20 – x) customers from his rivals. Each week the average shopper buys goods whose wholesale value is £A. Show that with a mark-up of x% the supermarket will have an anticipated weekly profit of
£ Ax/100 ((F+20k) –kx)
The only thing I don’t understand here is the 1/100. I get that: total profit= profit per customer x number of customers = (A + Ax –A) (F + k(20-x)) = Ax (F + 20k –kx).
But I don’t get where the 1/100 comes in, I know its cos ‘it’s a percentage’ but I don’t see where to insert it, mathematically.

Show that the manager can increase his profit by reducing his mark-up below 20% provided that 20k>F.

Here I differentiated weekly profit = f(x) with f(x) = 1/100 (AFx + 20Akx –AK(x)^2) to get
f’(x) = 1/100 (AF + 20Ak – 2Akx) > 0
AF + 20Ak – 2Akx > 0
We know A > 0 so we can divide both sides by A; F + 20k > 2kx
We know k > 0 so we can divide both side by 2k; x< (F + 20k)/2k
ð x> -(F + 20k)/2k
x must be smaller than 20 (given, mark-up below 20%)
ð -(F+20k)2k < 20
ð 40k > -F -20k
ð -F< 60k
ð ?
Can someone show me where I went wrong?

2. ok here's some more work

3. ok i figured out the percentage part, i.e. the first part of the question.
in the second part i can prove the statement if i can prove that F+20k < 40k
i already have F=20k> 2kx and x<20 so i have to prove 2kx<(F+20k)<40k