College Board’s explanation sucks…can someone try and explain this problem better?
Here’s a picture of the question; I’d rather not type it to avoid confusion of the signs and such:
http://oi57.tinypic.com/2n7dpmx.jpg
I reached an intermittent step but can’t really continue from there; can someone explain from the beginning?
@FutureDoctor2028 difference of squares or expanding and simplify:
(x+y)^2 - (x-y)^2 = (x+y + x-y)(x+y - (x-y)) = (2x)(2y) = 4xy >= 25
Dividing by 4, we have xy >= 25/4.
We are given 0 <= x <= y, so the least possible value of y occurs when x = y = 5/2 (note that if y was some value less than 5/2, then xy <= y^2 < 25/4, a contradiction).
@MITer94: Thank you so much! I stopped at xy>= 25/4 and didn’t know what to do from there!
Makes sense that x=y to be the least value since (5/2)(5/2) is 25/4, which is the least it can be. Thanks again.
@FutureDoctor2028 - I found this problem easier to do with a graph -
First, yes, simplify the larger equation as @MITer94 says.
Then, graph y >= (25/4) / x . the base graph is a hyperbola and a couple of points will clarify what it should look like. (you are constrained to the first quadrant by the other requirement)
also graph the required y >= x i.e. the y = x line and shade in the top portion in the first quadrant.
You will see that the region that satisfies both criteria is the top one, with a pointy part at its bottom. This pointy part is the point of intersection, which solves to be the point (5/2, 5/2)
In @MITer94’s solution it’s not as clear why the least possible value of y should occur at 5/2, but the reason is that any smaller, and that’s less than the square root of 25/4, so the x pops up to be bigger.
This problem seems hard for the SAT-I math, honestly.
@fretfulmother well, if you wanted a more rigorous solution, we have that
x <= y —> xy <= y^2 (multiplying both sides by y, where y >= 0).
Then 25/4 <= xy <= y^2, so y >= 5/2 (y is non-negative). Because y = 5/2 is attainable, it is the minimum possible value.
I do agree that it’s a little hard for SAT math since a lot of other techniques like plugging in, etc. won’t work as well. But I don’t think that the difficulty of this question is unreasonable.
@MITer94 - I didn’t say that you couldn’t be more rigorous, just that it might not be clear to the OP why 5/2 was smallest. 
BTW is '94 when you graduated MIT? We may have overlapped.
@fretfulmother no - '94 is my birth year, I am a junior at MIT and graduate in 2016.
(edit) Did you go to MIT?
@MITer94 - ah, makes sense!
@fretfulmother Thanks for the help! Yeah it is a hard question; its classified as a level 5 question from the answer key (from a difficulty level of 1-5).
@MITer94 - yes, graduated '93. Sorry, I didn’t see your update to the post!
…and, I couldn’t resist the siren song “confusing math problem please help” LOL
I thought I’d point out that you can use the info that y must be greater than or equal to x to conclude that the lowest value of y will occur when x=y. Then from the beginning you can substitute x=y to get rid of all the xs. Also couldn’t pass up “confusing math problem please help”!
^I don’t think you can jump from y>=x to x=y until you have already done the work to get to the 4xy=25 stage. At that point, you see that x and y play “symmetrical” roles. You wouldn’t have known that before you worked the problem (though it is in fact true this time).
But it certainly is a nice SAT problem. And notice that I did not jump in to claim that this one is easier without algebra!
Yeah, minimizing y doesn’t work in general.
@MITer94- I assumed that you were at MIT, but everytime I see you post I think of this: http://www.pubandshopsigns.com/pictures/159.jpg
(High St. Oxford)
