# Optimisation problem

1. May 26, 2007

### steven10137

1. The problem statement, all variables and given/known data
A right angled triangle with sides of length a, b and 2r is trapped within a semi-circle of fixed radius r. Given that all three vertices are on the semi-circle, find the values of a and b that will maximise the area of the triangle

http://img100.imageshack.us/img100/1804/areaoftrainglebl3.th.jpg [Broken]

2. Relevant equations
A(circle) = pi r ^2
A(triangle) using hero's formula (sqrt(s(s-a)(s-b)(s-c))) where s=(a+b+c)/2
OR
A(triangle)= (b x h)/2

3. The attempt at a solution
I am familiar with optimisation skills but am really stumped with this question, given that it is a scalene triangle.
If someone could give me a pointer in the right direction as to what formulas to use or attempt to derive, it would be appreciated
cheers

Last edited by a moderator: May 2, 2017
2. May 26, 2007

### phoenixthoth

3. May 26, 2007

### TheoMcCloskey

Hint - Since the vertices are always on the semi-circle, then the triangle will always be a right-triangle -- area of triangle will be (1/2) a * b

4. May 26, 2007

### steven10137

ahhhh

that is so obvious, cant believe i didn't figure that out.
I suppose I just didn't visualise the traingle in that way OR notice that it actually told me that it was right-angled ...

cheers
i shall now attempt to optimise it
thanks fellas

5. May 26, 2007

### steven10137

ok

so i have found that the area of the triangle is (ab)/2
Now i want to optimise this so that the area is maximised, so we need another equation.

I know that the area of the half-circle is (pi x r^2)
how can i now derive some other equation to maximise a and b?

6. May 26, 2007

### TheoMcCloskey

There is a relation - What can you say about the sides of a right triangle?

7. May 26, 2007

### steven10137

pythagoras ....
a^2 + b^2 = (2r)^2

take that in terms of 'a' or 'b' and then replace than in the area equation, find the derivate and then optain max/min .... cheers

sorry not thinking very well tonight ... thankyou for you patience.