Video Transcript
A farmer can improve the quality of his produce if he uses at least 18 units of nitrogen-based compounds and at least six units of phosphate compounds. He can use two types of fertilizers: A and B. The cost and contents of each fertilizer are shown in the table. Given that the graph represents the constraints in this situation, find the lowest cost the farmer can pay for fertilizer while providing sufficient amounts of both compounds.
Okay, in our table, we see the two types of fertilizers, A and B. Each one provides a certain number of nitrogen-based compounds per kilogram and a certain number of phosphate compounds per kilogram. Along with this, we’re told the cost in units of Egyptian pounds for one kilogram of each fertilizer type. Along with our table, we have a graph that shows us the constraints involved in this scenario. Our goal is to find the lowest cost that the farmer can pay for fertilizer while still providing sufficient amounts of these two compounds.
Knowing this, let’s clear away all of our text except for our first sentence. This sentence is helpful because it tells us what our constraints are that are depicted graphically. As we know, our goal is to minimize the total cost that our farmer spends on fertilizer. Knowing the types of fertilizer involved and the cost per kilogram of each one, we can say that the total cost the farmer will pay will be equal to 170 times the number of kilograms of fertilizer type 𝐴 plus 120 times the number of kilograms of fertilizer type 𝐵. This, then, is our objective function. This is the function we want to minimize, given our constraints.
As we’ve seen, those constraints are depicted in our graph. For example, the fact that our fertilizer needs to have at least 18 units of nitrogen-based compounds tells us that three times 𝐴 plus six times 𝐵 must be greater than or equal to 18. From our table, we note that each kilogram of fertilizer type A provides three units of nitrogen-based compounds, while each kilogram of type B provides six. So that’s one constraint. A second is that our fertilizer needs to have at least six units of phosphate compounds. We can write that as two times 𝐴 plus 𝐵 is greater than or equal to six. This expression comes from the fact that each kilogram of type A fertilizer provides two units of phosphate compounds, while each kilogram of type B provides one.
This inequality we’ve developed for the phosphate compounds involved can be seen on our graph using this pink line. Likewise, our first constraint can be shown using the second pink line. Because both of these constraints say that the total number of kilograms of A and B must be greater than or equal to some number, when we look at our graph, it’s actually the region marked out in gray which is our allowed region. Any points in this gray-shaded space would meet these conditions for the fertilizer quality. So when we look for the minimum cost that will satisfy these requirements, the points that we’ll need to investigate are points that border the great region in our graph.
There are three such points that we could think of as vertices: one right here, one here, and one here. Note, by the way, that neither negative 𝑥- nor negative 𝑦-values on our graph would meet our conditions because a negative value in either direction would indicate some negative number of kilograms of fertilizer type. All that said, our focus is in the first quadrant and on the area shaded in gray. Of all of these points, though, and there are infinitely many in the gray area, it’s only those three points we’ve highlighted in orange that might minimize our objective function.
Let’s look then at the coordinates of these three points. Looking closely at our top point, we see it has an 𝑥-value of zero and a 𝑦-value of six. The next point down from that has coordinates two, two. Our third point has coordinates six, zero. So then we’ll substitute each one of these three points into our cost function. And then we’ll consider the outputs of those calculations. For our first point, we have a cost of 170 times zero plus 120 times six; that’s 720. For our second point, the cost is 170 times two plus 120 times two or 580. And lastly, 170 times six plus 120 times zero is 1020.
Remembering that we want to minimize the total cost of our fertilizer, it’s the lowest of these three values that we’ll choose. So the smallest amount of money the farmer can spend on fertilizer while still satisfying these quality requirements is 580 Egyptian pounds.
