Re: Maxima Minima question

Hey numpty.

Hint: We know that a triangles area can be written in terms of A = 0.5*b*h. Can you use Pythagoras' Theorem to get h in terms of the other sides of the triangle (besides the base that is)?

Re: Maxima Minima question

you may have to walk me through this as i am still struggling

Re: Maxima Minima question

Given the three lengths, one is the base (b) and the height (h) is related to the two other sides.

The area of a triangle with lengths a and b and angle x in the middle is:

A = 1/2*a*b*sin(x) which you need to maximize.

You have a + b + c = 30 (the three sides) and you can get the angle using cos(x) = [b^2 + a^2 - c^2]/[2ab]

Now you have to eliminate the variables to get a maximum for A with respect to another variable (i.e. a,b, or c) using normal calculus.

Re: Maxima Minima question

Please check put if you have missed any part of the question that is some information about the type of triangle, isosceles / eqilateral.

Re: Maxima Minima question

I think it might be easier to use Heron's formula for the area of a triangle:

$\displaystyle A=\sqrt{(s)(s-a)(s-b)(s-c)}$

Fixing a and viewing b as a variable gives $\displaystyle A=\sqrt{(s)(s-a)(s-b)(s-(2s-a-b))}=\sqrt{(s)(s-a)(s-b)(a+b-s))}$

Differentiating with respect to b gives $\displaystyle \frac{dA}{db}=\frac{(s)(s-a)}{2\sqrt{(s)(s-a)(s-b)(a+b-s))}}(2s-2b-a)$, and since the first factor can not possibly be zero, $\displaystyle 2s-2b-a=c-b=0$, so $\displaystyle b=c$. Therefore the triangle in question is isosceles.

Setting $\displaystyle c=b$ and viewing a as a variable gives $\displaystyle A=\sqrt{(s)(s-a)(\frac{a}{2})(\frac{a}{2})}$, and differentiating gives $\displaystyle \frac{dA}{da}=-\frac{(s)(a)(2s-3a)}{4\sqrt{(s)(a^2)(s-a)}}$, so $\displaystyle 2s-3a=0$ and the triangle is equilateral.

I don't think the result is particularly surprising, but it's nice to see a proof.

- Hollywood