# Pre-Cal

posted by on .

I've been trying to work this problem out for the past week and still havent managed to get to the answer.

jeans unlimited sold 6000 pairs of jeans in october at an average of 44\$ each. The company president decides that she must increase the company's profits by increasing the price of a pair of jeans. Market research indicates that sales will drop 200 pairs for every 1\$ increase in price. If the cost of making a pair of jeans is 25\$ and all jeans made are sold , what price will maximize the profit?

The answer is 49.50\$ but im not sure how to get there. I know that i somehow have to find a quadratic equation and find the derivative of that to find the max which will essentially give me enough information to find the final answer. PLEASE HELP.

• Pre-Cal - ,

Profit=NumberJeans*price-cost
but numberjeans=6000-200(Price-44)

profit=(6000-200(P-44))P-25
I assume you know how to find max.
dp/dPrice=0=(6000-200(P-44))+P(-200)(1)

solve for P

0=6000-200P+8800-200P
400P=14,800
P= 47 dollars. Check my work.
I recommend on your graphing calculator graph the profit function
Profit vs P