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Max Area of isosceles triangle with perimeter 1

GreenGoblin

Member
Feb 22, 2012
68
Given this seemingly simple problem of maximising the area of an isosceles triangle with perimeter equal to 1. What is the best approach and how will I find a result easiest (I know how to get the answer but I need to be able to do these problems fast, so please help me look for a quick method if there is one?)

What I do is go for
max $(\frac {y}{2}^{2}) \sqrt(x^{2}-(\frac{y}{2})^{2})$
subject to $2x + y = 1$

Then rearrange y = 1 - 2x and substitute, maximise using the derivative etc but it takes absolute ages...
Is there a quicker way? How should I approach this kind of problem (a different problem may be given) in a timed scenario?

I got to $\frac{1-3x}{2\sqrt (x-\frac{1}{4}}$ which would make at 0 x=1/3 = y? This can't be right as it would make it an equilateral not isosceles. I also make the area $\frac{1}{12\sqrt (3)}$ which doesnt feel right.
 
Last edited:

CaptainBlack

Well-known member
Jan 26, 2012
890
Given this seemingly simple problem of maximising the area of an isosceles triangle with perimeter equal to 1. What is the best approach and how will I find a result easiest (I know how to get the answer but I need to be able to do these problems fast, so please help me look for a quick method if there is one?)

What I do is go for
max $(\frac {y}{2}^{2}) \sqrt(x^{2}-(\frac{y}{2})^{2})$
subject to $2x + y = 1$

Then rearrange y = 1 - 2x and substitute, maximise using the derivative etc but it takes absolute ages...
Is there a quicker way? How should I approach this kind of problem (a different problem may be given) in a timed scenario?

I got to $\frac{1-3x}{2\sqrt (x-\frac{1}{4}}$ which would make at 0 x=1/3 = y? This can't be right as it would make it an equilateral not isosceles. I also make the area $\frac{1}{12\sqrt (3)}$ which doesnt feel right.
Might I suggest you use Heron's formula for the area:

\[A^2=s(s-a)(s-b)(s-c)\]

Where \(s\) is the semi-perimeter and \(a, b,\) and \(c\) are the sides.



CB
 
Last edited:

GreenGoblin

Member
Feb 22, 2012
68
I basically did, albeit unintentionally. I have the same expression it gives for the area.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,488
max $(\frac {y}{2}^{2}) \sqrt(x^{2}-(\frac{y}{2})^{2})$
subject to $2x + y = 1$
Shouldn't this be $\max\frac {y}{2} \sqrt{x^{2}-\left(\frac{y}{2}\right)^{2}}$ (i.e., the first y shouldn't be squared)? Otherwise, you are measuring area in $\mbox{cm}^3$...
 

GreenGoblin

Member
Feb 22, 2012
68
You are correct,
I typed that up wrong,

I used the correct version as you stated in my calculations.
I can't see at all where I've gone wrong but I know my answer must be wrong... but that is the only extreme point I can find. It looks more like a minimum if anything at all but I cna't find another.

Just to clarify, I did that, subbed in y = 1-2x, differentiated and set to 0. What I put is what I got. Forgive me not typing up all my notes as it would take a couple hours with all that tex. This goes againsy my instrinct though and I think there should be a simpler way.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Might I suggest you use Heron's formula for the area:

\[A^2=s(s-a)(s-b)(s-c)\]

Where \(s\) is the semi-perimeter and \(a, b,\) and \(c\) are the sides.



CB
I basically did, albeit unintentionally. I have the same expression it gives for the area.
So as we have an isoscellese triangle let a be the non-equal side, then the sides are \(a, (1-a)/2, (1-a)/2\) so you need to maximise:

\[ A^2=(1/2)\left(\frac{1}{2}-a\right)\left(\frac{a}{2}\right)^2\]

CB
 

GreenGoblin

Member
Feb 22, 2012
68
So as we have an isoscellese triangle let a be the non-equal side, then the sides are \(a, (1-a)/2, (1-a)/2\) so you need to maximise:

\[ A^2=(1/2)\left(\frac{1}{2}-a\right)\left(\frac{a}{2}\right)^2\]

CB
Yes, that's the same. It's just a rearranged form and skipping the substitution stage. Your a is my x, no difference.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Yes, that's the same. It's just a rearranged form and skipping the substitution stage. Your a is my x, no difference.
Note finding the side that minimises \(A^2\) also minimises \(A\), so you do not need to take the square root.

CB
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Given this seemingly simple problem of maximising the area of an isosceles triangle with perimeter equal to 1. What is the best approach and how will I find a result easiest (I know how to get the answer but I need to be able to do these problems fast, so please help me look for a quick method if there is one?)

What I do is go for
max $(\frac {y}{2}^{2}) \sqrt(x^{2}-(\frac{y}{2})^{2})$
subject to $2x + y = 1$

Then rearrange y = 1 - 2x and substitute, maximise using the derivative etc but it takes absolute ages...
Is there a quicker way? How should I approach this kind of problem (a different problem may be given) in a timed scenario?

I got to $\frac{1-3x}{2\sqrt (x-\frac{1}{4}}$ which would make at 0 x=1/3 = y? This can't be right as it would make it an equilateral not isosceles. I also make the area $\frac{1}{12\sqrt (3)}$ which doesnt feel right.
An equilateral triangle is an isosceles triangle (and an equilateral triangle is the answer, it does maximise the area of all isosceles triangles of perimeter 1)

CB
 

GreenGoblin

Member
Feb 22, 2012
68
My answer is correct then, I haven't done anything wrong.
 

CaptainBlack

Well-known member
Jan 26, 2012
890

loquetedigo

New member
May 12, 2015
14

MarkFL

Administrator
Staff member
Feb 24, 2012
13,735
< 0.0481125224 => TrianCal
Yes, I believe it was already established clearly in this thread that the maximum area of an isosceles triangle having a perimeter of 1 is given by:

\(\displaystyle A_{\max}=\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{3}\sin\left(60^{\circ}\right)=\frac{\sqrt{3}}{36}\)

As stated, for a given perimeter, the isosceles triangle having the greatest area will be equilateral.

Additional information posted in threads is certainly appreciated, but please ensure you are adding information and not simply reiterating what has already been given.