Expert: Paul Klarreich - 1/10/2008

Question
1) find the dimensions of the trapezoid of greatest area that can be inscribed in a semicircle of radius R.

>> i am completely lost on this sort of questions because they don't give us numbers to work with.. all i got down were the two area equations for the trapezoid (5/2)(a+b) and semicircle (pie*r^2/2). i'm not even sure if i'm using the right equations either...

thank you for you help

Answer
Hi, Alex,

I know -- things like this strike terror into the hearts of many studnts. (Along with problems starting with the dreaded words "Show that").

Try this trick:  Put in some number, like:

find the dimensions of the trapezoid of greatest area that can be inscribed in a semicircle of radius 73.  [An awful number, but that's as it should be.]

Try to solve the problem, such as:

Assume the semicircle is centered at the origin and has the equation:

x^2 + y^2 = 73^2

The base of the trapezoid will go from  (-73,0) to (73,0) and the top from (-x,y) to (x,y).

The area formula is h/2(a + b), so that will be:

A = y/2(2x + 2(73))  << Try NOT to do arithmetic with the 73. Don't write 146.

Now  y = sqrt(73^2 - x^2) and we can substitute.

Are you getting the idea, now?  Perhaps we can start putting R's in now --


A = y/2(2x + 2R)

y = sqrt(R^2 - x^2)

The rest of it is pretty standard.  I'll leave it to you, and suggest you look at

http://en.allexperts.com/q/Calculus-2063/2008/1/Maximum-minimum-problem-8.htm

for a useful trick.