Interpreting the supremum

Question

I came across this percentage problem in a high-school textbook:

The marked price of a toy is $\$ 40$. What is the largest possible discount per cent that can be given such that the selling price is not less than $\$36 $?

And the solution given is:

Let $x\%$ be the discount percent. $$40(1-x\%) \geq 36$$ $$x\% \leq 10\%$$ $\therefore$ The largest possible discount per cent is $10\%$.

It is of course perfect. As a teacher, I start to think of what my students would answer. One possible answer I am interested in is this:

Let $y\%$ be the largest discount percent.

If we still keep the inequality sign, i.e. $40(1-y\%) \geq 36$, what are we actually calculating? Are we considering the supremum of $y$? Is it even sensible to keep the inequality sign?

To rephrase my question, how should we interpret the answer, i.e. $y\% \leq 10\%$? I try to work on that but it is kind of difficult for me to give a clear explanation. Any help would be appreciated. If you think the question is not clear, please tell me so that I can amend it.

[ Please find my argument in the answer below. ]

Answer 1

Strictly speaking, the maximum discount, let $y\%$, satisfies the following inequality: $$y\%\geq x\%\text{, for every possible discount }x\%$$ Yet, since $y\%$ is still a discount as this is discussed in our problem, it should also satisfy the same restrictions as $x\%$ does, so, I feel that the inequality sign should remain.

One can see - of course not a student, but for extension purposes - the abovementioned problem as follows:

Find the maximizing value of $x$ for the function: $$f(x)=x$$ under the conditions: $$\begin{align*} 40(1-x)-36&\geq0\\ -x&\leq0\\ x&\leq10 \end{align*}$$

Under that phrasing, since the set of the restrictions is conmpact and $f$ is continuous, this problem has a solution and, the notion of supremum, due to compactness, coincides with that of maximum, so it is "well-hidden", Moreover, in this case the problem is made trivial, since the set of restrictions is a singleton; $\{10\}$.

So, back to the initial question, I think that there is no reason to drop the inequality sign since the maximum discount $y\%$ is, well, still a discount! :)

Answer 2

If we still keep the inequality sign, i.e. $40(1-y\%) \geq 36$, what are we actually calculating?

Not calculating anything here. Just stating the premise of the problem: $\,y\,$ is a discount (doesn't matter at this point whether it's the largest or not), and so $\,y\,$ must satisfy the given inequality.

Is it even sensible to keep the inequality sign?

Not only sensible, but in fact required.

Imagine for example that you changed the problem to require "the largest possible discount per cent which is a multiple of 3% that can be given such that the selling price is not less than $36$". In that case the inequality $\,y\% \le 10\%\,$ stays the same, but the actual answer will be $\,y\% = 9\%\,$.

To rephrase my question, how should we interpret the answer, i.e. $y\% \leq 10\%$?

In plain English: the maximum discount $\,y\,$ cannot be larger than $\,10\%\,$.

Note that at this point $\,10\%\,$ is only an upper bound, not yet proved to be an actual maximum. It happens that $\,10\%\,$ is actually the answer in this case, but that hasn't been established, yet.

I try to work on that but it is kind of difficult for me to give a clear explanation.

The missing step, I think, is to recognize that, in this case, any percent $\,0\% \le x\% \le 10\%\,$ qualifies as an eligible discount according to the stated rules. Therefore the maximum discount must be the largest amount all numbers in the range $\,[0\%, 10\%]\,$, and that number is $\,10\%\,$.

Answer 3

The largest discount percentage that can be applied to a \$$40$ toy to ensure that the selling price is not less than \$$36$ is the discount price which will make the selling price exactly equal to \$$36$. If we call this discount percentage $y$ then we can drop the inequality sign and obtain the following equation:

$$40(1−y/100)=36$$

If we solve this equation we obtain $y=10$.

Now if we call $x$ a discount percentage that can be applied to a \$$40$ toy to ensure that the selling price is not less than \$$36$ then the equation above turns into an inequality, since there are an infinite number of possible answers:

$$40(1−x/100)≥36$$

If we solve this equation we obtain $x≤10$. In this case $y$ is simply the supremum of this infinite set:

$y=\text{sup}\{x\in\mathbb{R}|x≤10\}=10$

Therefore we have an equality for $y$, since $y$ can only take on a single value. Although $y$ also satisfies the inequality $y≤10$ that is not the same thing as being defined in terms of that inequality, in which case $y$ would have an infinite number of values, which is false. Furthermore, $y$ satisfying the inequality $y≤10$ provides us with less information about $y$ than the equality $y=10$, so there is no reason to keep the inequality.