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Video: Solving Story Problems with Nonlinear Inequalities

Tim Burnham

Using the skills we have learned in other videos on nonlinear inequalities, we learn how to apply them in order to construct and solve suitable nonlinear inequalities from written descriptions of problems.

13:03

Video Transcript

In this video, we’re gonna write and solve an inequality which is nonlinear in order to answer a question raised in a story problem or a real-life scenario. We’re gonna have to carefully define our variables and think about the information that we’ve been given and how it translates into equations or inequalities. Okay, let’s take a look.

A packaging company needs to create some trays without lids. They will cut out square corners 𝑥 inches by 𝑥 inches from square sheets of card which is ten inches by ten inches. The area of the base must be at least forty-eight square inches and the tray must be at least one inch deep. Write and use an inequality to find the range of values that 𝑥 can take.

So we’ve been given a diagram and we can see that they got these sides here. They’re gonna be folded up just to make her a little tray. It’ll look something like that. So the height of the tray is gonna be equal to 𝑥 cause those sides are just gonna fold up without any sort of lid, and those would be some sort of glue going on the edges around here and here and here in order to stick those together.

So it’s also- it’s gonna be a square base, so this distance is gonna be the same as this distance here. But what will that distance be equal to? Well, if the whole width of the piece of card was ten inches and we’ve taken off 𝑥 on this side and 𝑥 on this side, the bit that’s left in the middle there is ten take away 𝑥 take away another 𝑥. And that is ten take away two 𝑥. So really the first tip for these sort of questions is do a diagram; it just helps you to think about the problem and get your head around what all the different variables and bits of information mean.

So let’s think about some of the criteria that we’ve got in the question here. We-we’ve been told that the area of the base must be at least forty-eight square inches. Well, the area is basically the kind of- the width times the length of the base or ten minus two 𝑥 times ten minus two 𝑥. And when we multiply that out, we’ve got ten times ten is a hundred. Ten times negative two 𝑥 is negative twenty 𝑥. Negative two 𝑥 times ten is negative twenty 𝑥. And negative two 𝑥 times negative two 𝑥 is positive four 𝑥 squared.

So tidying that up, we’ve got four 𝑥 squared minus forty 𝑥 plus a hundred is an expression for the area of the base. It’s probably worth just writing down a proper definition of what 𝑥 is at this time. We were given 𝑥 in the equation, but let’s just make sure that we’re absolutely clear in our mind what 𝑥 represents. 𝑥 represents the number of inches that we’re gonna be cutting from the corner of the cards, so it’s the length of this bit, or this bit, et cetera on that card. So the area of the base in terms of 𝑥 is four 𝑥 squared minus forty 𝑥 plus a hundred. And as we said, the question tells us that that’s gotta be at least forty-eight square inches, so its allowed to be equal to forty-eight square inches or it’s allowed to be bigger. So the area of the base is greater than or equal to forty-eight, and four 𝑥 squared minus forty 𝑥 plus a hundred is greater than or equal to forty-eight. So we’ve written an inequality. Now, we’ve just got to use it to find the range of values that 𝑥 can take.

But before we go on and try and solve the problem, there are a couple of other little inequalities that I’m gonna write down first. We’re told that the tray must be at least one inch deep, so this means that the 𝑥 value must be at least one; so it could be greater than or equal to one. But also if you think about it, if 𝑥 was bigger than five, then we’ll be cutting over halfway across that card. The whole thing is not gonna work. If 𝑥 was equal to five, then we would have the zero area base. So 𝑥 actually probably needs to be less than five as well. So this is just the physical constraint of the problem; if 𝑥 was five or above, then we wouldn’t have a tray at all. And this was another part of another constraint that was given to us in the question because the tray has to be at least one inch deep. So we can represent that as one individual inequality since it’s a continuous region. And that would be one is less than or equal to 𝑥 is less than five. So we’re actually gonna use that second inequality here, the most that’s gonna give us the most information, but we also need to bear in mind the first one because they’re important constraints from the question as well. So let’s go ahead and try and solve four 𝑥 squared minus forty 𝑥 plus a hundred is greater than or equal to forty-eight.

Well, the first thing that I notice is that all of those numbers are divisible by four, so I’m gonna divide three by four just to make the numbers a bit easier. So that means that 𝑥 squared minus ten 𝑥 plus twenty-five is greater than or equal to twelve. Well, now I’m gonna subtract twelve from both sides, so we’ve got an inequality which is greater than or equal to zero cause that’s gonna help us when we draw a graph and we’re trying to interpret the region. So we’ve got a quadratic inequality. Now, if we find the roots of 𝑥 squared minus ten 𝑥 plus thirteen, that will tell us where the sort of 𝑦-graph- where the 𝑦-coordinate of the graph would be zero, so numbers where it cuts the 𝑥-axis. We can then see which 𝑥-coordinates generate 𝑦-coordinates above the 𝑥-axis and which-which generate coordinates below the 𝑥-axis, and we can solve this inequality in that way. So let’s just look at the parameters on that quadratic equation. We’ve got a positive coefficient of 𝑥 squared; that’s one 𝑥 squared. So because that’s positive, it’s a nice happy quadratic curve. So that’s what it’s gonna look like. It would — If we were putting 𝑦 equals 𝑥 squared minus ten 𝑥 plus thirteen, that would cut the 𝑦-axis at 𝑦 equals thirteen. So now we need to know where would 𝑦 equals 𝑥 squared minus ten 𝑥 plus thirteen cut the 𝑥-axis. And to do that, we need to find the roots of the equation. And looking at that, it doesn’t look like it’s gonna factorise very easily, so we’re gonna have to use either completing the square — In fact, I’m gonna use the quadratic equation. Now remember, that is for an equation of the type 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero where 𝑎 is not zero, 𝑥 would be equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎 . So 𝑎 is the coefficient of 𝑥 squared, 𝑏 is the coefficient of 𝑥, and 𝑐 is the number on its own. So to find out where the- this particular quadratic 𝑦 equals all of that would cut the 𝑥 axis, we’re gonna put 𝑦 equal to zero, so that whole thing would be equal to zero. So it’s gonna be in this particular format here.

So we already said that 𝑎 is equal to one and 𝑐 is equal to thirteen, so we just need to know now that 𝑏 is equal to negative ten, and we can plug those numbers into the quadratic formula. So to find the roots of 𝑦 equals 𝑥 squared minus ten 𝑥 plus thirteen, we’re basically plugging those numbers in here. And we’ve got 𝑥 is equal to the negative of negative ten plus or minus the square root of negative ten squared minus four times one times thirteen all over two times 𝑎, which is two times one. And that simplifies down to ten plus or minus root forty-eight over two. And of course root forty-eight simplifies to four root three. And ten and four on that numerator there have a common factor of two, so I can factor that out which will enable me to cancel down the two on the numerator and the denominator, which just leaves me with five plus or minus two root three. So our two solutions for when 𝑥 squared minus ten 𝑥 plus thirteen is equal to zero are 𝑥 is five plus two root three, which is about eight point four six to two decimal places, or when 𝑥 is equal to five minus two root three which is one point five four to two decimal places. So we know where it cuts the 𝑥-axis. We know it was a happy curve. We know where it cuts the 𝑦-axis. We can now go ahead and sketch the curve.

So transferring that information onto a sketch, we’ve got our axes here. We know it cuts the 𝑦-axis at thirteen, so that’s up here. We know it cuts the 𝑥-axis sort of say here and here. And since it’s a symmetric parabola, we know that the midpoint will be at five because you’ve got where it cuts the 𝑥-axis at five minus two root three and five plus two root three. So the curve is gonna look something like this, coming down to here and then going up symmetrically on the other side. So what does 𝑦 equal 𝑥 squared minus ten 𝑥 plus thirteen actually represent? Well, it’s not the area of the base of a tray. Remember, the area of the base was four 𝑥 squared minus forty 𝑥 plus a hundred. But what it will do if we look at this, if we look at the 𝑥-coordinates that generate 𝑦-coordinates greater than or equal to zero, that’ll tell us valid values of 𝑥 to solve our original problem involving the area of the base being greater than or equal to forty-eight square inches. So we’ve got to find the 𝑥-values on this graph that generate 𝑦-coordinates of greater than or equal to zero, so that’s these regions up here. So an 𝑥-value of five minus two root three, that’s included cause that does generate a 𝑦-value of zero using this equation. And anything smaller than that will also generate correct values for 𝑦. And five plus two root three will generate a 𝑦-value of zero and anything to the right of that will as well. So we’ve got two regions of valid values for 𝑥 from this criterion of our problem.

So let’s represent those solutions on a number line like this. I’ve just got the 𝑥-values from zero to ten. A value of five minus two root three, that’s one point five four, or less will give us a valid region. And values of eight point four six to two decimal places, five plus two root three, or more will also give us valid values of 𝑥 for the area constraint. Now let’s consider those other physical constraints that we had, you know: you can’t cut the card- you can’t cut the corners too big otherwise you wouldn’t have a tray, and the tray has to be at least one inch deep. So let’s put that onto a line- number line as well. And it’s allowed to be equal to one, but it’s got to be less than five, so it’s gotta be in this region here. So whereabouts do those two regions overlap? Well, if we draw a line from one — a dotted line coming down here and dotted line coming down here, this region here — both things are true. So over here, this criterion is met, but this one isn’t. And over here, this criterion is met, but this one isn’t. So really it’s just this very small region here of 𝑥-values that we’re allowed to use. So it’s allowed to be one, and it’s allowed to be five minus two root three cause that’s a solid dot here and that’s a solid dot here, and it’s allowed to be anything in between.

So the final answer is the valid values of 𝑥, in order to meet all the criteria that we had for the tray in terms of the area of the base and the depth of the tray and fit to be a proper physical tray, are 𝑥 is greater than or equal to one and 𝑥 is less than or equal to five minus two root three. Any value of 𝑥 within that range will give us a tray which matches all the criteria.

So let’s just summarise the process that we’ve been through then. Firstly, we had to read the question really carefully and we drew diagrams which helped us to visualise what was going on and to really understand the problem. Second, we had to clearly define any variables that we were gonna be using in the question. Then we had to construct inequalities from the criteria given; some were sort of explicitly given in the question and some we had to sort of implicitly work out, like the fact that if we took a cut of more than five inches, then we wouldn’t actually be making a tray. And lastly, we had to solve the inequalities. In fact, in our case, we were able to simplify the inequality a bit, and we generated a sort of a corresponding inequality which we then solved that which gave us the same 𝑥 solutions to our original problem. And lastly, you need to check your answer’s sensible and it meets all the criteria given in the question.