Algebra 2nd Edition: Chapter 4 Pg. 91

[fahzia]

I am having difficulty understanding the answer to the question on Pg 91. The answer is on pg. 97

question: -4<a<4 and -2<b<-1
answer:

max of a: LT(4)
min of a: GT(-4)

max of b: LT(-1)
min of b: GT(-2)

ab most +ve: GT(-4)x GT(-2)= LT(+8)
ab most -ve: GT(-2)x GT(4)= GT(-8)

Where is GT(4) in the ab most -ve coming from? GT(-2) is the most -ve value of b I understand but the most +ve value of a is LT(4) and not GT(4). I guess I don't fully understand how the answers are calculated.

[jen]

Hi there,

The short answer is that b could be -2 and a could be 4, thus maling ab = -8.

But the larger point of problems like this is that, in order to calculate the maximum and minimum values of ab, you CANNOT just take the "smallest a times the smallest b" and the "largest a times the largest b." You have to be more strategic.

So, if -4 < a < 4 and -2 < b < -1, then we are going to calculate using extreme values for a (-4 and 4) and extreme values for b (-2 and -1). But wait! There is also a ZERO in the range for a, so we'll need to check that out as well. So, if I asked you the min/max for a + b, or b^2 - a, or ab, or any crazy thing, you would consider the following:

For a, -4, 0, and 4
For b, -2 and -1

(I know a is actually "greater than" -4 and "less than" 4 -- we'll just phrase our final answer with inequalities also to take that into account.)

So, for min and max ab, our options are:
(-4)(-2) = 8
(-4)(-1) = 4
(0)(-2) = 0
(0)(-1) = 0
(4)(-2) = -8
(4)(-1) = -4

Since -8 and 8 are the smallest and largest, those are our answers. (Some people are able to do this faster by strategizing ahead of time instead of trying all the possibilities, as I did above. But don't forget zero if it's in a range! Zero didn't affect the answer to this problem, but certainly would affect the answer to similar problems of this type.)

Sincerely,
Jen

[fahzia]

Thanks for the explanation Jen. I think you missed the -ve sign in the case of b i.e. it is 2 < b < -1 but that would not change the answer.

[jen]

You're right -- I edited my original explanation above so others visiting the page will see it correctly. Thanks!