Video: Using Graphs of Rational Functions to Solve Real-World Problems

Consider the square prism shown in the diagram. Write its surface-area-to-volume ratio in terms of 𝑥. Give your answer in standard form. The diagram shows the graph of the surface-area-to-volume ratio of the prism as a function of 𝑥. Which of the following is an approximate value of 𝑥 for which the surface-area-to-volume ratio is 1? [A] 6 [B] 3.3 [C] 2.3 [D] 1.3 [E] 1.5.

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Video Transcript

Consider the square prism shown in the diagram. Write its surface-area-to-volume ratio in terms of 𝑥. Give your answer in standard form.

And there’s also a second part to this question which we’ll deal with later. It’s quite clear that to answer this question, we’re going to need to form expressions for both the surface area and the volume of the square prism. Remember, the surface area of a three-dimensional shape is the combined area of all its faces. We could say that our square prism is a cuboid. And cuboids have six rectangular, or sometimes square, faces. So, it’s really sensible to be methodical about how we calculate each of their areas.

Let’s begin by forming expressions for the area of the rectangle at the front of our diagram. The formula for area of a rectangle is its base multiplied by its height. So, the area of this rectangle can be found by multiplying one plus two 𝑥 by one plus 𝑥.

Now spot that we’ve been asked to give our answer in standard form. This is not to be confused with the standard form that we often use to describe numbers which are very very small or very very large. When we write an expression in standard form, it’s important that each variable has one factor. So, we can’t have any brackets in there. And we write these in descending powers. So, in this example, we’re going to write it in descending powers of 𝑥.

So, lets expand these brackets as normal. We could use the grid method or the FOIL. method. Let’s look at the FOIL. method for expanding these brackets. F stands for first. We multiply the first term in the first bracket by the first term in the second bracket. One multiplied by one is one. O stands for outer. We multiply the outer term in each bracket. That gives us 𝑥. I stands for inner. We multiply the inner terms. And two 𝑥 multiplied by one is just two 𝑥. And L stands for last. We multiply the last term in each bracket. Two 𝑥 multiplied by 𝑥 is two 𝑥 squared.

And all that’s left to do is to simplify this expression by collecting like terms and write it in descending powers of 𝑥. That’s from largest to smallest. 𝑥 plus two 𝑥 is three 𝑥. So, we can write the expression for the area of this rectangular face as to 𝑥 squared plus three 𝑥 plus one.

But this is a prism. So, this means the area of the face on the opposite side, that’s kind of the back of the prism, if you will, is the same. And, in fact, since the prism has a square base, we can see that the rectangular face to the right of our shape also has an area of two 𝑥 squared plus three 𝑥 plus one as does the face directly opposite that. That’s the face on the left-hand side of our prism.

This means we have four faces with an area of two 𝑥 squared plus three 𝑥 plus one. And we can expand these brackets by multiplying each term by four. And we see that the area of the four rectangular faces is eight 𝑥 squared plus 12𝑥 plus four.

And next we have the face at the top of our prism. It has a base of one plus two 𝑥 and a height of one plus two 𝑥. So, its area is one plus two 𝑥 all squared, or one plus two 𝑥 multiplied by one plus two 𝑥. And expanding as we did before, or distributing these brackets using the FOIL method, we get one plus two 𝑥 plus two 𝑥 plus four 𝑥 squared, which in standard form is four 𝑥 squared plus four 𝑥 plus one.

But how many other faces have this area? Well, we have an identical face at the bottom of our prism. So, there are two faces with an area of four 𝑥 squared plus four 𝑥 plus one. And distributing these brackets by multiplying each term by two, we get the total area of these two faces to be eight 𝑥 squared plus eight 𝑥 plus two.

Remember, we said that the surface area is the combined area of all six faces on our square prism. So, that’s eight 𝑥 squared plus 12𝑥 plus four plus eight 𝑥 squared plus eight 𝑥 plus two. And since we’re only adding and adding is commutative, we can see that we don’t really need these brackets. In descending powers of 𝑥, we have eight 𝑥 squared plus eight 𝑥 squared, which is 16𝑥 squared, 12𝑥 plus eight 𝑥 is 20𝑥, and four plus two is six. So, the total surface area of our square prism is 16𝑥 squared plus 20𝑥 plus six.

Our next step is to form an expression for the volume of the square prism. The volume of a prism is found by multiplying the area of its cross-section by its length. Now we said that the cross section of this prism is a square. So, that’s its base multiplied by its height multiplied by its length. And in fact, we can generalise that the volume for any cuboid is its base multiplied by its height multiplied by its length.

And in any order, the volume of this square prism is one plus two 𝑥 multiplied by one plus two 𝑥 multiplied by one plus 𝑥. Remember, multiplication is commutative. So, it doesn’t really matter if we put one plus 𝑥 at the front or in the middle. This is going to give us the same answer when we distribute these brackets. And to distribute these brackets, we begin by expanding, or distributing, any two of the brackets. Let’s choose one plus two 𝑥 multiplied by one plus two 𝑥. We already saw that that gives us four 𝑥 squared plus four 𝑥 plus one.

And to multiply this by one plus 𝑥, the grid method could be a nice way to check that you don’t make any mistakes. We begin by multiplying four 𝑥 squared by one. That’s four 𝑥 squared. Four 𝑥 multiplied by one is four 𝑥. And one multiplied by one is one. Then four 𝑥 squared multiplied by 𝑥 is four 𝑥 cubed. Four 𝑥 multiplied by 𝑥 is four 𝑥 squared. And one multiplied by 𝑥 is 𝑥. We have four 𝑥 cubed plus eight 𝑥 squared, that’s four 𝑥 squared plus four 𝑥 squared, plus five 𝑥 plus one.

And all that’s left is to write the ratio of the surface area to its volume. To achieve this, we divide the expression we formed for the surface area by the expression we formed for the volume of the square prism. it’s 16𝑥 squared plus 20𝑥 plus six all over four 𝑥 cubed plus eight 𝑥 squared plus five 𝑥 plus one. Let’s clear some space and consider the second part of this question.

The diagram shows the graph of the surface-area-to-volume ratio of the prism as a function of 𝑥. Which of the following is an approximate value of 𝑥 for which the surface-area-to-volume ratio is one? Is it A) six, B) 3.3, C) 2.3, D) 1.3, or E) 1.5?

And it’s important to note that your answers might be in a different order. So, how do we answer this question? Well, we’re told that the surface-area-to-volume ratio is one. And if we look at the graph, we can see that the surface-area-to-volume ratio of the prism is represented by the values on the 𝑦-axis. The surface-area-to-volume ratio is one is over here. So, we’re going to add a horizontal line onto our graph until we hit the curve. From there, we’re going to add a vertical line downwards until we hit the 𝑥-axis. To work out the value of 𝑥, which corresponds to a surface-area-to-volume ratio of one, we need to find the scale on the 𝑥-axis.

We can see that five small squares on our 𝑥-axis represent one unit. If we divide by five, we then see that one square is equal to one-fifth of a unit, or 0.2 of a unit. Our value of 𝑥 lies halfway through a square though, so we divide again by two. And we see that half a square is equal to 0.1 of a unit. Our value of 𝑥 is one and a half squares above the number three. So, that’s three plus the value of one small square, which is 0.2, plus the value of that half a square, which is 0.1, which is 3.3.

And we see that the approximate value of 𝑥 for which the surface-area-to-volume ratio is one is 3.3. And we can check this if we want. We check it by substituting it into the ratio we formed in the first part of this question. Substituting 𝑥 equals 3.3 into that formula gives us 0.991, which is extremely close to one, the ratio we were looking for.

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