Question

A manufacturer of a certain brand of appliance estimates that he can sell 5000 units a year at Php 900.00 each and that he can sell 1500 units more per year for each Php 100.00 decrease in price? What price per unit will give the greatest returns?

Answer 1

Let x be the number of price decreases. Then the revenue is

r(x) = (900-100x)(5000+1500x)
= 100*500(9-x)(10+3x)
= 50000(-3x^2+17x+90)

Now find when dr/dx=0 and you can figure the price for maximum revenue.

Answer 2

let the number of 100 decreases be n

number sold = 5000 + 1500n
selling price per item = 900 - 100n

return = (5000+1500n)(900-100n)
= 500(10 + 3n)(100)(9 - n)
= 50000(90 + 17n - 3n^2)
d(return)/dn = 50000(17 - 6n)
= 0 for a max
6n = 17
n = 17/6 = appr 2.8

assuming we can only decrease in multiples of 100
for 2 decreases:
number = 8000
price = 700
return = 5 600 000

for 3 decreases:
number = 9500
price = 600
return = 5 700 000

there should be 3 decreases of 100