Topic : Calculus in business
Degree : 42
Department : Mechanical Engineering
Syndicate : C
Intro of calculus
RATE OF CHANGE
BEHAVIOUR OF
GRAPH
Application of
calculus in
business
Maximize profit
Minimize loss
Terms of business
Cost of production 𝑐 𝑥
Revenue 𝑅 𝑥
Profit 𝑃 𝑥 = 𝑅 𝑥 − 𝑐 𝑥
T-shirt's factory
◦ We sell x number of units at p$/t-shirt.
◦ Demand equation
p = 10 − 0.001𝑥
So, our revenue will be
𝑅 𝑥 =𝑥×𝑝
Where p is selling cost
𝑅 𝑥 = 𝑥 10 − 0.001𝑥
Maximizing the revenue
◦ We have to differentiate and find the maximum value for revenue
𝑅′ 𝑥 = 10 − 0.002𝑥
◦ To find maxima
10 − 0.002𝑥 = 0
𝑥 = 5000
selling cost will be
𝑝 = 10 − 0.001 5000
𝑝 = 5$
And revenue will be
𝑅 𝑥 = 5000 10 − 0.001𝘹5000
𝑅 𝑥 = 25000$
Maximizing the profit
◦ Our cost of production is
𝐶 𝑥 = 5000 + 2𝑥
◦ Our profit equation will be
𝑃 𝑥 =𝑅 𝑥 −𝑐 𝑥
𝑝 𝑥 = 10𝑥 − 0.001𝑥 − (5000 + 2𝑥)
To maximize our profit we have to differentiate the profit equation and find
maximum value
𝑝′ 𝑥 = −0.001𝑥 2 + 8𝑥 − 5000
𝑥 = 4000 𝑢𝑛𝑖𝑡𝑠
So we have to sell 4000 t shirts to maximize our profit and that profit will be
𝑝 4000 = 10 × 4000 − 0.001 × 4000 − (5000 + 2 × 4000)
𝑝 4000 = 11,000$
Thank you