Topic : Calculus in business Presenter : Muhammad Hassan Masood 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