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: Simplify the constraints<br />### Divide the first inequality by 40: $x + 2y \ge 58$<br />### Divide the second inequality by 10: $4x + 3y \ge 132$<br />### Divide the third inequality by 40: $3x + y \ge 54$<br /><br />## Step 4: Determine the objective function<br />### The objective is to minimize the operating cost: $C = 70x + 40y$<br /><br />## Step 5: Find the vertices of the feasible region<br />### The feasible region is defined by the intersection of the constraints. We find the vertices by solving the system of equations formed by pairs of boundary lines.<br />### Intersection of $x + 2y = 58$ and $3x + y = 54$: Solving this system gives $x = 10$ and $y = 24$.<br />### Intersection of $4x + 3y = 132$ and $3x + y = 54$: Solving this system gives $x = 18$ and $y = 18$.<br />### Intersection of $4x + 3y = 132$ and $x=0$: Solving this system gives $x=0$ and $y=44$.<br />### Intersection of $x+2y=58$ and $y=0$: Solving this system gives $x=58$ and $y=0$.<br />### Since $x=0, y=44$ and $x=58, y=0$ are outside the feasible region defined by $3x+y \ge 54$, we discard them.<br /><br />## Step 6: Evaluate the objective function at each vertex<br />### At $(10, 24)$: $C = 70(10) + 40(24) = 700 + 960 = 1660$<br />### At $(18, 18)$: $C = 70(18) + 40(18) = 1260 + 720 = 1980$<br /><br />## Step 7: Determine the minimum cost<br />### The minimum cost is $1660, which occurs when $x = 10$ and $y = 24$.