# An airline has three types of airplanes and has contracted with a tour group to provide...

## Question:

An airline has three types of airplanes and has contracted with a tour group to provide transportation for a minimum of 76 first-class, 54 tourist-class, and 120 economy-class passengers. The first plane costs $4100 for the trip and can accommodate 44 first-class, 10 tourist-class, and 26 economy-class passengers; the second plane costs$4600 for the trip and can accommodate 14 first-class, 16 tourist-class, and 42 economy-class passengers; the third plane costs \$5900 for the trip and can accommodate 28 first-class, 24 tourist-class, and 14 economy-class passengers. How many of each type of airplane should be used to minimize the operating cost?

## Minimum cost:

We can solve the following problem by a clever inspection to the options. By realizing that no three-combinations of planes can accommodate, we can start optimizing the solution into four-combinations.

First let's call the first plane as {eq}A {/eq}, the second plane as {eq}B {/eq} and the third plane as {eq}C {/eq}. Then let's denote the first class as {eq}F {/eq}, the tourist class as {eq}T {/eq} and the economy class as {eq}E {/eq}. So if we say {eq}120 E {/eq}, it means 120 passengers for the economy class.

By inspection, we can see that it is not possible to accommodate the minimum requirements by either 3 combinations of {eq}A, B, C {/eq}.

Minimum A B C
76 44 14 28
54 10 16 24
120 26 42 14

To see that, let's start by looking at the requirement for economy class, that is, {eq}120E {/eq}. Notice that {eq}(2A + B)E < 120E {/eq} as well as {eq}(2C + B)E < 120E{/eq}. By adding all the combination, we'll find that {eq}(A + B + C)E < 120E{/eq}. The only option to fully accommodate the economy class is to have {eq}(3B)E = 3(42)= 126E {/eq}. However, by doing so, we will be lacking

$$76F - 3(14)F = 34F \\ 54 - 3(16)T = 6T \\$$

By this, we conclude that there is no possible 3-combinations to accommodate the minimum requirements. Now, we have to prove that 4-combinations is the optimal solution. To do so, we will show that five combinations of the cheapest plane will be greater than 4-combinations of either planes, {eq}5A > 4(A,B,C) {/eq}. Since there is no reason for us to choose {eq}4C {/eq}, we can safely prove that our claim is correct by showing that

$$5A > 3B+C \\ 5(4100) > 3(4600) + 5900 \\ 20500 > 19700$$

Our goal is to find for the minimum cost of 4-combinations. Again, we will look at {eq}120E {/eq}. By the previous findings, we lack {eq}34F, 6T {/eq} by having {eq}3B {/eq}. Then, choosing the cheapest option to complete the requirements, we now have one candidate, {eq}3B + A {/eq}. Next, since we want to minimize the cost, we want to maximize the number of {eq}A {/eq} as much as possible. Notice that {eq}(3A+B)E = 120E {/eq}. By this combination, we will find that we lack

$$54T - 3(10)T-(16)T = 8T$$

Hence, this combination will not be possible. Then, by following the logic that we want to maximize the number of {eq}A {/eq} and {eq}B {/eq} as much as possible, we'll then try {eq}2B + 2A {/eq}. However, by doing this we will be lacking

$$54T - 2(10)T-2(16)T = 2T$$

Now, our next option will be {eq}3B + A {/eq} but recall that {eq}3B + A {/eq} is already our candidate. Hence, 3 second plane and 1 first plane is the minimum.