A tour company has two types of airplanes.the T3 and the S5, and contracts requiring accommodations for a minimum of 2320 first-class, 1320 tourist-class, and 2160 economy-class passengers. The T3 costs 70 per mile to operate and can accommodate 40 first-class, 40 tourist-class, and 120 economy-class passengers, whereas the S5 costs 40 per mile to operate and can accommodate 80 first-class, 30 tourist-class and 40 economy-class passengers How many of each type of airplane should be used in order to minimize the operating cost? The tour company should use square T3 airplanes and square T5 airplanes to minimize the cost. (Type whole numbers.)

## Step 1: Define Variables<br />### Let $x$ be the number of T3 airplanes and $y$ be the number of S5 airplanes.<br /><br />## Step 2: Set up the constraints<br />### First-class: $40x + 80y \ge 2320$<br />### Tourist-class: $40x + 30y \ge 1320$<br />### Economy-class: $120x + 40y \ge 2160$<br />### Non-negativity: $x \ge 0$, $y \ge 0$<br /><br />## Step 3: Define the objective function<br />### The objective is to minimize the operating cost: $C = 70x + 40y$<br /><br />## Step 4: Simplify the constraints<br />### Divide the first-class constraint by 40: $x + 2y \ge 58$<br />### Divide the tourist-class constraint by 10: $4x + 3y \ge 132$<br />### Divide the economy-class constraint by 40: $3x + y \ge 54$<br /><br />## Step 5: Graph the constraints and find the feasible region<br />### The feasible region is the area where all constraints are satisfied. The vertices of the feasible region are the points where the constraint lines intersect.<br /><br />## Step 6: Find the vertices of the feasible region<br />### Intersection of $x + 2y = 58$ and $4x + 3y = 132$: Solving this system gives $x = 18$ and $y = 20$.<br />### Intersection of $4x + 3y = 132$ and $3x + y = 54$: Solving this system gives $x = 12$ and $y = 18$.<br />### Intersection of $3x + y = 54$ and $x + 2y = 58$: Solving this system gives $x=10$ and $y=24$.<br /><br /><br />## Step 7: Evaluate the objective function at each vertex<br />### At $(18, 20)$: $C = 70(18) + 40(20) = 1260 + 800 = 2060$<br />### At $(12, 18)$: $C = 70(12) + 40(18) = 840 + 720 = 1560$<br />### At $(10,24)$: $C = 70(10) + 40(24) = 700 + 960 = 1660$<br /><br />## Step 8: Determine the minimum cost<br />### The minimum cost is $1560$ at $(12, 18)$.