Sum is from application of calculus in commerce and economics Share with your friends Share 0 Ankita Agarwal answered this Dear Student, Selling price of 1 item =Rs.(330−x)(Given) Selling price of x items =Rs. x(330−x) Cost price of x items =Rs.(x2+10x−12) Profit = Selling price − Cost price ⇒ Profit =(330x−x2)−(x2+10x−12) =320x−2x2+12 Now, P=−2x2+320x+12 Differentiating above equation, we get dPdx=−4x+320 Differentiating againg w.r.t. x, we get d2Pdx2=−4⇒ Always negative To maximize the profit, dpdx=0−4x+320=0x=−320-4=80 ∵d2Pdx2<0 for any value of x, thus 80 items must be sold so that his profit is maximum. Regards 0 View Full Answer