Answer:
Selling price that will maximize profit is $57
Explanation:
Given;
Costs to manufacture and distribute a backpack = $14
Number sold, n = [tex]\frac{2}{x-14}+5(100-x)[/tex]
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 = [tex][\frac{2}{x-14}+5(100-x)]\times(x-14)[/tex]
or
P = 2 + 5(100 - x)(x - 14)
or
P = 2 + 5(100x - 1400 - x² + 14x)
differentiating with respect to x
we get
[tex]\frac{dP}{dx}[/tex] = 0 + 5(100 - 0 - 2x + 14)
or
[tex]\frac{dP}{dx}[/tex] = 5(114 - 2x)
put
[tex]\frac{dP}{dx}[/tex] = 0 for point of maxima or minima
5(114 - 2x) = 0
or
114 - 2x = 0
or
x = $57
Now,
[tex]\frac{d^2P}{dx^2}[/tex] = 5(0 - 2) = -10
[hence, negative result means x = 57 is point of maxima]
Therefore,
Selling price that will maximize profit is $57