Given ranges of two variables, what is the range of their product?

Question

A GRE prep question asks, "If $-13 < a < -2$ and $1 < b < 9$, which of the following could be equal to the product of $a$ and $b$?"

Potential answers: $-20$, $-18$, $-15$, $-14$, $-13$, $-9$.

The provided explanation is to multiply the inequalities,

\begin{align} (-13)(1) &<& ab &<& (-2)(9) \\ -13 &>& ab &>& -18 \\ \end{align}

which gives the solutions $-15$ and $-14$ only.

But it seems to me that $a = -10, b = 2, ab = -20$ works as well, but is excluded by their inequality.

I wasn't confident in their method of multiplying inequalities, but this answer suggests that it is indeed valid.

What am I missing?

Answer 1

if $-13 < a < -2$ and $1 < b < 9$,

then $$-13(9) < ab < -2(1)$$

$$-117 < ab < -2$$

Hence it is possible to obtain all those valuves.

• you should be expecting a negative answer.

• also, in terms of magnitude, $2 < |a|< 13$ and $1 < |b| < 9$, hence $1(2) < |ab|< 9(13)$.

Answer 2

The test is wrong. The cited question was for positive ranges.

We have $-13 < a < -2$. To make postive

$2 < |a| < 13$.

And we have

$1 < b < 9$

It is a basic axiom that if $x < y$ and $k > 0$ then $xk < yk$. As a direct result we have if $x < y$ and $k < 0$ then $xk > yk$. (Because $k < 0\implies $k - k < 0 - k\implies 0 < -k$. So $x(-k) < y(-k)$ and $xk > yk$.)

And therefore if $0 < x < y$ and $0 < u < v$ then $xu < yu$. And $uy < vy$ so $xu < yv$.

So $2*1 < |a|b= -ab < 9*13$. And because "negatives flip signs"

$-9*-13 < ab < -2$.

... so what if you had renges that were positive AND negative?

Do it in steps.

$-2 < a < 5$ and $-1 < b <3$

Case 1) $a \ge 0; b \ge 0$

$0\le a < 5$ and $0\le b< 3$ so $0\le ab < 15$.

Case 2) $a < 0; b < 0$ then

$-2 < a < 0; -1 < b < 0$ so $0< ab < 2$.

Case 3) $a < 0; b\ge 0$ then

$0 < -a < 2; 0 \le b < 3$ and $0 < -ab < 6$ and $-6 < ab < 0$

Case 4) $a \ge0; b< 0$ then

$0\le a < 5; 0 < -b < 1$ so $0 < ab < 5$ and $-5< ab < 0$.

Combining all 4 cases

$-6 < ab < 15$

=====

Um... $a = -4$ and $b = 5$ is certainly possible.

The link given assumes that $a,b,c,d\ge 0$ which is not the case here.

We have $-13 < a < -2$ and $b > 0$ so $-13b < ab < -2b$.

And $1 < b < 9$ and $-13 < 0$ so $-117 < -13*9 < -13b < -13$ and $-18 < -2*9 < -2$.

So $-117 < ab < -2$ are all possible.

So the test is dead wrong.

Sad thing is... this happens.

.... but come on!

$ab = -20$ can happen if $a=-4; b= 5$

$ab = −18$ can happen if $a=-3;b =6$

$ab = −15$ can happen if $a=-3; b=5$

$ab = −14$ can happen if $a = -3; b= 4\frac 23$

$ab = −13$ well, let $a= -\sqrt 13$ (which is between $-3$ and $-4$) and $b = \sqrt 13$

Answer 3

One more answer:

1)$ 1<b<9;$

2) $-13 <a<-2$.

Consider $ba$: Multiplying the negative number $a$ by a positive number $b$ makes it smaller.

1) Now pick the upper positive bound for $b,$ i.e. $9$, and multiply with the lower negative bound for $a$, i.e. $-13$, to get the lower bound for ab, hence $(-13)9 < ab$.

2) Similarly pick the negative upper bound for $a$, i.e. $-2$, and multiply by the positive lower bound for $b$, i.e. $1$, to get the upper bound for $ab$, i.e. $ab< -2(1)$,

3) Altogether:

$(-13)9< ab<-2(1)$.

Check for numbers in the above range.

Answer 4

Alternatively, draw the feasible region $a\times b=(-13,-2)\times (1,9)$ and calculate the product $(z(a,b)=ab)$ at the corner points: $$z(-13,9)=-117, z(-2,1)=-2, z(-13,1)=-13, z(-2,9)=-18.$$ Since the function $z(a,b)$ is continuous and concave, its range is $(-117,-2)$ on the given domain.