Expert: Ahmed Salami - 11/1/2008

Question
The manager of a supermarket usually adds a mark-up 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.

I have no idea why the problem is applied to differentiation...

Anyway, here are the questions.
1. Show that with a mark-up of x% the supermarket will have an anticipated weekly profit of £1/100*Ax((F+20k)-kx)

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

Answer
Hi Veda,
The manager has a loyal core of F customers i.e those who'll always buy, whatever the price.
By lowering the markup to x%, he has extra k(20-x) customers.
The total number of customers he then has is,
N = F + k(20 - x)
 = F + 20k - kx
If the average shopper buys goods whose wholesale value is £A weekly, the profit he makes from one shopper is
p = £A.x%
 = £A.x/100
1) The anticipated weekly profit is therefore,
P = Np
 = (F + 20k - kx).£A.x/100
 = £(1/100)Ax(F + 20k - kx)

2)From P = £(1/100)Ax(F + 20k - kx)
P = £(A/100)x(F + 20k - kx)
 = £(A/100)(Fx + 20kx - kx^2)
To find the maximum value of the profit, we find the point where dP/dx = 0
For y = ax^n, dy/dx = n.ax^(n-1)
And so,
dP/dx = £(A/100)(F + 20k - 2kx)
at dP/dx = 0
£(A/100)(F + 20k - 2kx) = 0
F + 20k - 2kx = 0
F + 20k = 2kx
x = (F + 20k)/2k

Now, from the information we've had that the manager attracts an extra k(20 - x) customers when the markup is x,
we can see that x should be below 20% otherwise (20 - x) would be negative and he's then not attracting extra
customers to make more profit.
Going back to the point of maximum profit
x = (F + 20k)/2k
When x < 20
(F + 20k)/2k < 20
(F + 20k) < 40k
F < 20k
as required.

Regards.