Cristina S. answered • 03/24/17
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Hank, Thank you for your comments, I completely understand your explanation of "small". Certainly, I didn't look at the date, so I didn't realize it was posted 10 days ago when I originally posted my comment. I am revising my comments and I hope you will do the same. Also, the reason I didn't post a complete solution, is because I posted a comment, not because I couldn't provide a solution. Thank you for your comments and I'm sorry if I upset you.
Hank L.
Cristina, with all due respect, the only person here who is wrong is you.
First, your solution never answers the problem. As you have it right now, it only says X is on the interval of 0 to 7, which is ok, but what are the pairs of numbers? If x is 1, then (x-7) is -6 and your product is -6. If x is 2, then (x-7) is -5 and your product is -10. So clearly, x can't just be anything on the interval of 0 to 7
Second, as Ken pointed out, this was solved over a week ago. You will note that I never contested Ken's answer by defending mine. This is because upon re-examination of the question, I thought his answer was probably more in line with what an algebra 2 class would ask and deemed it a non-issue, as the original poster never had any follow up questions.
Third, I stand by my answer. Here is the original question verbatim: One number is 7 less than a second number. Find a pair of such number that their product is as small as possible. Ok at issue here is the word "small". What is small?
When I think of "small" I think of numbers like 0.00002 or 1 x 10-14. As such, I don't view -5.9 x 1016 to be a small number. I think of it as a very large negative number. If I said I had a small quantity of apples, I would think of one or two apples, not -25 apples. So my interpretation of the problem was as I answered it. I could see a high school teacher throwing in a trick question like that.
Had the question been tagged with calculus instead of algebra 2, I might have instinctively thought differently and arrived at the conclusion that Ken had by taking the first derivative of x(x-7) and setting it equal to zero, thus finding the critical point at x = 7/2. However, like I said, I interpreted the word "small" differently. In physics, positive and negative is simply a direction that a vector points. The magnitude is what is "small" or "large". I would never call a large negative number "small", but maybe in math you do, and that's why I didn't argue my point with Ken. Hope this makes sense!
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03/24/17
Hank L.
Hi Cristina, I'd edit it out, but it appears that I can't edit comments. You might have meant to just post a comment, but your original "comment" is actually posted as an answer to the question. You can edit those (I can still go in and edit my March 14 answer below) but I can't go in and edit or delete the above comment. Maybe there's a time limit statute on edits for comments like 2 hrs or something, whereas answers stay open indefinitely (or maybe until the original poster selects a favorite answer). But just so you know, you are listed as an "answer"... in fact you appear as the "latest answer" to this question. Right now there are three answers (you being the third, so you might as well finish the solution :) ) Sorry I can't edit the comment (you can see for yourself, you probably can't edit the comment you made on Ken's answer earlier today). It won't matter much, it's an old question and the only people looking at it right now are you, me, Kenneth, and maybe the original asker :) Take care!
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03/24/17
Cristina S.
Hank, no problem, I don't really care about the post, but thanks for trying :) Ok, I will finish the solution, since you asked and no one is watching, it only makes sense for you to ask I'd finish and for me to actually finish it :) Since you entertained me with negative apples, I'll try to find a solution to match it, I'll give it my best shot, although who can compete with negative apples? Just so you know, this is not the solution I had in mind.
Here it goes: Since I am looking at the product x(x-7) and I want it to be as "small" as possible, whatever that means, no one seems to care about proper definitions anymore, I visualize |x| and |x-7| as the measurements of the sides of a rectangular to be able to use the area of the rectangle as product of
|x|*|x-7|.
Here it goes: Since I am looking at the product x(x-7) and I want it to be as "small" as possible, whatever that means, no one seems to care about proper definitions anymore, I visualize |x| and |x-7| as the measurements of the sides of a rectangular to be able to use the area of the rectangle as product of
|x|*|x-7|.
For the reasons I stated before, 0<x<7, so x>0, so I don't have to use absolute value for side x of my rectangular.
Since x-7<0 when 0<x<7, I will use 7-x as my other side of the rectangle. Now both sides of my imaginary rectangle have positive values, I can work with them.
In order for x(x-7) to be as "small" as possible, I want it to be as hugely negative as possible, and therefore the area of my rectangle A=x(7-x) has to be as enormous as possible, that is both sides have to be huge! If both sides are huge, they are so close in size that
their difference is 0. So,
(7-x)-(x)=0, or 7-2x=0, or 2x=7, therefore x=7/2. It follows that the original "small" product is:
x(x-7)=7/2*(7/2-7)=7/2*(-7/2)= -49/4 which is such a "small " product, that is as "small as possible"
The pair is: (7/2, -7/2).
I hope I could prove to you that I can find the "as small as possible" product on the planet :) even "smaller" than your -25 apples!
Have a great weekend!
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03/24/17
Kenneth S.
03/24/17