Geometry

[SilverSprite]

Could anybody show me how to do this the correct way? I got 112 but sort of a stupid way to do it and time consuming.

In the rectangles in the diagram, the lengths of AB, BC, AH, HG are integers. The area of rectangle ABIH is 6 and the area of rectangle IDEF is 15. What is the largest possible area for rectangle ACEG?

[Kalkewl8ter]

Oh fun...Pascal question...
Ok, your answer is right; this is a possible solution:
You want the side lengths of rectangle ACEG to be as large as possible. To do this, the perimeters of rectangles ABIH and IDEF must be as large as possible. Therefore you set AB=6, AH=1, DE=15, and ID=1. The maximum area of rectangle ACEG is (6+1)*(15+1)=112.

[bugzpodder]

You want the side lengths of rectangle ACEG to be as large as possible. To do this, the perimeters of rectangles ABIH and IDEF must be as large as possible.

thats not very good reasoning to me. but i couldn't think of anything better.

[A$$Bandit]

Sounds perfectly *reasonable* to me

[bugzpodder]

if the question didn't say intergral lengths, then largest area will occur on the largest perimeter. it only so happens when they tell u that its intergral lengths

[Kalkewl8ter]

First line of the question, bugz:

"In the rectangles in the diagram, the lengths of AB, BC, AH, HG are integers. "

[bugzpodder]

right i noticed that

[SilverSprite]

if you read up kalk he said if they werent integral. sheesh

[bugzpodder]

lol leave her along sprite stop arguing over pointless things!

[Kalkewl8ter]

Oops! I can't read...oh well. Let Sprite be that way...we're all used to it anyways.

[sql_lall]

AB = x
BI = 6/x
DE = 15/y
ID = y

All integers:
=> x = 1,2,3 or 6
=> y = 1,2,3 or 15

Area = 6 + 15 + x(15/y) + y(6/x)
= 21 + 15(x/y) + 6(y/x)
=> better to have large x/y, not large y/x
=> best when x = 6, y = 1
Area = 21 + 15*6 + 1 = 112

[bugzpodder]

=> better to have large x/y, not large y/x

what if the question is 15(x/y)+14.99999(y/x)?? is it better always to have larger x/y?? no! maybe the maximum of x/y is 5 and the maximum of y/x is 15! also there are other possibilities depending on the coefficients of (x/y) and (y/x) such that we need to have a ratio of 5:3 for example to attain the maximum (given that you can only use a few values for x and a few values for y)! of course when there are no restritions for x/y then your observations is correct because x/y can be infinitely large

[sql_lall]

Ok then, as there are only 16 possibilities, you can check them all, and still get 112.

However, let x/y = p
=> 21 + 15(x/y) + 6(y/x)
= 21 + 15p + 6(1/p)
= 21 + 6(p+1/p) + 9p

1) if p<1, 1/p > p
=> 21 + 6(p+1/p) + 9(1/p) > 21 + 6(p+1/p) + 9(p)
=> There is a larger solution with 1/p, and 1/p > 1

only have to worry about p >1
2) as p (>1) increases, p +1/p increases, 9p increases
=> maximum value of 21 + 15p + 6(1/p) is when p is maximum, => x/y is maximum
=> x = 6, y = 1
which is just what i was saying before

[bugzpodder]

no offense or anything, i think your arguement is still poor and for some parts i was unable to follow it

[sql_lall]

Don't worry, no offense taken. I kinda am getting used to people not understanding my explanations

Anyway, the answer is 112

(Just curious, you said:
"also there are other possibilities depending on the coefficients of (x/y) and (y/x) such that we need to have a ratio of 5:3 for example to attain the maximum",
and i was just wondering if you had actaul values in mind)


Oh, and Silver Sprite, you said you used a stupid, time consuming way to do it?? How did you do it??

[Kalkewl8ter]

He did it the same way that I did it, which may be weak, but definitely not time-consuming.

[SilverSprite]

Yes it is time consuming. I would'nt know if i had it rigth without trying some out. I wanted to know if there was a mathy way to go about doing it.

[bugzpodder]

i'll look at the problem (and you think i've looked at it already)