# Maxima Minima question

• Feb 28th 2013, 02:26 PM
numpty
Maxima Minima question
I'm new to the forum and i'm not even sure that this question should be posted in this section but i hope someone can point me in the right direction.

A piece of string 30 inches long has its two ends joined together and is stretched by three pegs so as to form a triangle.

What is the largest triangular area that can be enclosed by the string?

I think I need to express the Area of the triangle in terms of the length of string (which is my problem) then differentiate and equate to zero to find the maxima.

Help appreciated.
• Feb 28th 2013, 04:33 PM
chiro
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)?
• Mar 1st 2013, 03:20 AM
numpty
Re: Maxima Minima question
you may have to walk me through this as i am still struggling
• Mar 1st 2013, 04:17 PM
chiro
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.
• Mar 1st 2013, 08:46 PM
ibdutt
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.
• Mar 2nd 2013, 09:57 AM
hollywood
Re: Maxima Minima question
I think it might be easier to use Heron's formula for the area of a triangle:

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

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

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

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

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

- Hollywood