Discuss a real-world application of the use of calculus-based derivatives for maximizing profits.

Question

 Derivatives

Answer 1

a common place this would show up is when you are trying to determine the optimal price and quantity to produce for a product. you would end up looking for the spot where marginal revenue equals marginal cost in order to find the quantity that optimizes the profit function.

Answer 2

Demand function D(p) = -2p + 25

That is, the demand deceases by 2 when the price increase $1

Revenue function R(x) = x*D(x) = x(-2x+25) = -2x^2 + 25x

0 = dR/dx = R'(x) = -4x + 25


x = -25/-4 = 25/4 = 6.25

 

So the max revenue occurs when the price is $6.25.

 

The max revenue is then
-2(25/4)^2 + 25(25/4) = -2(625/16) + 625/4 = 625/4 - 625/8 = 625/8 = 78.125