Question

A charter airline finds that on its saturday flights from philadelphia to london, all 120 seats will be sold if the ticket price is $200. however, for each $3 increase in ticket price, the number of seats sold decreases by one. (a) find a formula for the number of seats sold s if the ticket price is p dollars. s(p) = (b) over a certain period, the number of seats sold for this flight ranged between 70 and 115. what was the corresponding range of ticket prices (in dollars)? (enter your answer using interval notation.)

Answer 1

`(a)\\ \text{A charter airline finds that on its saturday flights from philadelphia to london, }\\ \text{all 120 seats will be sold if the ticket price is }\$200.\\ \\ \text{However for each }\$3\text{ increase in ticket price, the number of seats sold }\\ \text{decrease by one.}\\ \\ \text{Let the number of seats sold is s, when the ticket price is p dollars.}\\ \\ \text{at price 200 the number of seats sold is 120, so one point is }(200, 120)`

`\text{and at the price of 203 the numebr of seats sold is 119, so }(203, 119)\\ \\ \text{hence the slope of linear model is}=(119-120)/(203-200)=-(1)/(3)\\ \\ \text{Hence the formula for the number of seats using point slope form is}\\ \\ s-120=-(1)/(3)(p-200)\\ \\ \Rightarrow s-120=-(1)/(3)p+(200)/(3)\\ \\ \Rightarrow s=-(1)/(3)p+(200)/(3)+120\\ \\ \Rightarrow s=-(1)/(3)p+(200+360)/(3)\\ \\ \Rightarrow s=-(1)/(3)p+(560)/(3)`

Hence the function is:
`s(p)=-(1)/(3)p+(560)/(3)`

`(b)\\ \text{over a certain period, the number of seats sold for this flight ranged between 70 and 115.}\\ \\ \text{so we have}\\ \\ 70\leq -(1)/(3)p+(560)/(3) \leq 115\\ \\ \Rightarrow 70\leq (1)/(3)(560-p) \leq 115\\ \\ \Rightarrow 210 \leq (560-p) \leq 345\\ \\ \Rightarrow 210-560 \leq -p \leq 345-560\\ \\ \Rightarrow -350 \leq -p \leq -215\\ \\ \Rightarrow 215\leq p\leq 350\\ \\ \text{hence the range of ticket prices in dollar is: }[215, \ 350]`