Scenario: I am interested in selling cookies in my online bakery store. But before that, I need to find out how much I should price it to maximize revenue. I surveyed 1500 people and found out that if I price it at $2.00 a cookie then 800 people would buy it. If I price it at a dollar more at $3.00. a cookie, then 600 people would buy it. On the other hand, if I price it at a dollar less at $1.00 a cookie, then 1000 people would buy it. Based on this information, I noticed that for every one dollar increase in price, 200 fewer people would buy it. 1. Solve the scenario problem above for the optimal item price that maximizesrevenue. Be sure to show all your work and reflect on your findings.

[tex]\begin{gathered} -200\text{ = }\frac{y\text{ - 800}}{x\text{ - 2}} \\ y\text{ - 800 = -200(x - 2)} \\ y\text{ - 800 = -200x + 400} \\ y\text{ = -200x + 400 + 800} \\ y\text{ = -200x + 1200} \end{gathered}[/tex]

Revenue = price X number of people

[tex]\begin{gathered} R(x)\text{ = x }\times\text{ y} \\ R(x)\text{ = x(-200x + 1200)} \\ R(x)=-200x^2\text{ + 1200x} \\ R^{\prime}(x)\text{ = -400x + 1200} \\ To\text{ maximize , R(x) = 0} \\ -400x\text{ + 1200 = 0} \\ 400x\text{ = 1200} \\ x\text{ = }\frac{1200}{400} \\ \text{x = 3} \end{gathered}[/tex]

Optimal item price = $3