Isosceles triangle and its maximum area

**totalnewbie** · February 24th 2006, 08:40 AM

I have to determine "a" so that the area of triangle is the largest. The lenght of slant side is "b".
Look at the picture and explain me where is the mistake.

**ThePerfectHacker** · February 24th 2006, 09:02 AM

You need to use the product rule when taking the derivative, also the chain rule.

The derivative of $\sqrt{b^2-\frac{a^2}{4}}$ is $-\frac{1}{4}a\left(b^2-\frac{a^2}{4}\right)^{-1/2}$ .
Thus,
$s'=\frac{1}{2}\left(b^2-\frac{a^2}{4}\right)^{1/2}-\frac{1}{8}a^2\left(b^2-\frac{a^2}{4}\right)^{-1/2}$ .
But $s'=0$ ,
Thus,
$\frac{1}{2}\left(b^2-\frac{a^2}{4}\right)^{1/2}=\frac{1}{8}a^2\left(b^2-\frac{a^2}{4}\right)^{-1/2}$
Since $\left(b^2-\frac{a^2}{4}\right)^{-1/2}\not =0$ divide both sides by it to get (also mutiply both sides by 8),
$4\left(b^2-\frac{a^2}{4}\right)=a^2$ .
Thus,
$4b^2-a^2=a^2$ ,
Thus,
$2b^2=a^2$
Thus,
$a=b\sqrt{2}$
Q.E.D.

**totalnewbie** · February 24th 2006, 11:31 AM

I got the same answer. The chain rule surprised me. I have not used it before. It took me some time to understand how it works.

**Rich B.** · February 24th 2006, 11:57 AM

Greetings:

Why do you believe yourself to be mistaken? Your result, i.e., a = b*sqrt(2) is correct. By the way, because your triangle is isosceles with base = (sqrt(2))(side), it is a 45-45-right triangle. This is sensible if you think of an isosceles triangle as one-half of a rhombus with side lengths, b. A bit of thought makes clear that the rhombus of greatest area is a square, half of which is an isosceles-right triangle -- and determined without the aid of Newton/Leibniz!

...for what it's worth.

Regards,

Rich B.

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