(A sly plot into getting one of you geeks to crack this question i cant seem to do)..

An isosceles triangle has sides of length x, x and p − 2x where p is
the length of the perimeter of the triangle. Find the value of x which
maximises the area of the triangle for fixed p and all the angles of the
triangle for this value of x.

I think I know the theory for how to get the answer, but the algebra is disgusting and doesnt work.

:respect:

Good luck folks.

Disgusting algebra, how bad can it be? You put it into the eqn for area of triangle and Bob's your uncle surely?

yeah i know that, lol...buh

1/2 base * height...

getting the height by pythag...

i get horrible algebra..

Is this GCSE or Alevel

neither really, it comes from an oxford entrance exam, however it is the last question on the maths part.

the only a level part of it is the differentiation to find maximized area.

I think using the area of the triangle as your objective function isn't the right way to go - as you say, it gets far too messy.

After you've found the 'height', k, through pythagoras, I suggest you try to maximise the area of the 'square', of sides k and p-2x. Finding the x which maximises the area of the square would be the same thing as finding the x that maximises the triangle. It'd obviously be a lot easier to work with too.

Didn't bother to go all the way through but a few tricks with the chain rule when differentiating, it seems ok. Setting the first order condition (FOC), you'll obtain a nice quadratic to solve for x. If you need to prove it's a maximum, taking the second order condition would get very messy so I'd just go with explaining through common sense - not that hard to do once you see what you get with the FOC.

Hooray

Someone else who knows it's maths and not math (unless you're American)

for K i get ... root [xp - (p^2)/4]

times that by (p-2x)/2 gives you the area right?
or your saying get the area of the square, in which case you times by what?

to differentiate this is horrible i think (or my diff skills suck), unless it works out lovely into a quadratic? or my K is wrong...

Wouldn't the maximum area be when the triangle is equilateral?

Listen to Syn.

LSE boy right there.

I got the same for K.

I'd times K by (p-2x) to get your objective function (the square). You could use (p/2 - x) which would directly give you the area of your triangle - it doesn't really matter... whether you use 2(p/2 - x) or (p-2x), it won't affect your value for x when differentiating because you could always divide through by 2 after taking the FOC.

So you try to find the value of x such that (p-2x)[xp - (p^2)/4]^(1/2) is maximised.

As I said, some 'tricks' with indices (writing x as (x^2)^(1/2) so you can include everything within the root sign) would make things easier. And instead of trying to get rid of the nasty dominator (after differentiating, you'd obviously end up with [....something...]^(-1/2), just think about what the condition needs when equating the expression to 0. If I didn't make any silly errors, I ended up with a quadratic 0=3x^2 -2x -p^3

...haven't tried completing the square to solve it. Since reasonable enough though.

You > Me

We need latex tbh :coffee:

As well as spoiler tags letters <_<

That's what I originally thought - makes sense intuitively. I'll see if I can come up with a counter-example.

The fact that the 3rd side is 'p − 2x' muddies the water somewhat and makes my head hurt.

thank you very very very much...

got it now...woooooooo :whacky: :whacky: :whacky:

you's a clever guy

you're right you know... :clap: :clap: :clap:

:woohoo:

Really? :blink:

Counter-example: let p=40.

For an equilateral triangle, x=13.3 (recurring).

The area of that would be less than if the sides were x=15, meaning 2p-x=10.

edit: I'm talking rubbish

:unsure:

:pal:

Yeah, scrap that...I'm talking bollocks with the counter-example.

:pal:


Is this making anyone else feel really stupid?

You < Me

My head hurts just reading about it.

Those who have to work it out. :pal:

No its really quite simple. I just cant be bothered with it as when I intially glanced at the question I thought it was a GSCE simultaneous equation type question so was not in the correct mind set to differentiate :good: :coffee:

:goodpost:

Two of my flatmates do math-related degrees and are always on about this stuff. It's alien to me.

ok i get it, you find the answer is p/3 which genius letters saw instantly, thus the angles being 60.


but what is still confusing me is this..

i'd times K by (p-2x) to get your objective function (the square). You could use (p/2 - x) which would directly give you the area of your triangle - it doesn't really matter... whether you use 2(p/2 - x) or (p-2x), it won't affect your value for x when differentiating because you could always divide through by 2 after taking the FOC.

what square, are you saying that the maximised area of the square - (sides x and p-2x) will give the same x for the triange, its just easier to deal with?

or u saying the area of the triangle is base * height, coz i thought it was half* base *height. in which case area would be height * (p-2x)/2

:unsure:

Think about it this way:

If you had to find the value of positive (or else this won't have a solution) x that maximises f(x) = x^2 - x^3, you'd find that f'(x) = 2x - 3x^2 ==(FOC)==> x = 2/3

But if you had to find the value of x that maximises 2f(x) = 2x^2 - 2x^3..........you'd find that f'(x) = 4x - 6x^2 ==(FOC)==> x = 2/3

The value of x stays the same because whether the objective function is mutiplied by a constant doesn't make a difference to your value of x. Because you'll just divide through by that constant in the FOC.

In the same way, if you were maximising the area of the triangle, you'd have (1/2)ab. Or for the whole equilateral triangle, you'd have ab. Or for the whole square, you'd have 2ab. Either way, it doesn't make a difference to your value of x.

saaaaffffffeeeeee mate....cheers ....appreciate the help.
:nanacheers: :bow:

:satan: