Answer:
Selling price that will maximize profit is $57
Step-by-step explanation:
Given;
Costs to manufacture and distribute a backpack = $14
Number sold, n =

here, x is the selling cost of the bag
Now,
Profit = Total revenue - Total cost
or
P = nx - 14n
or
P = n(x - 14)
or
P =
![[(2)/(x-14)+5(100-x)]*(x-14)](https://img.qammunity.org/2020/formulas/business/high-school/981vwwe3ohm91o5msexzfxpprm7mk9bdpw.png)
or
P = 2 + 5(100 - x)(x - 14)
or
P = 2 + 5(100x - 1400 - x² + 14x)
differentiating with respect to x
we get
= 0 + 5(100 - 0 - 2x + 14)
or
= 5(114 - 2x)
put
= 0 for point of maxima or minima
5(114 - 2x) = 0
or
114 - 2x = 0
or
x = $57
Now,
= 5(0 - 2) = -10
[hence, negative result means x = 57 is point of maxima]
Therefore,
Selling price that will maximize profit is $57