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#### 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.

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.

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