optimisation prob: find dimensions of rectangle that can be

[ku1005]

with the following optimisation problem im stuck, as im not sure how to get either y in terms of x or x in terms of y so that I can then do a MAx calculation, if someone would be able to show me how that would be great!

"Find the dimensions of the rectangle of largest area that can be inscribed in an equilteral triangle of side L, if one side of the rectangle lies on the base of the triangle"

I set up diagram with base of rectangle = x and height = y

therfore Area Rectangle = xy (this is what we are required to Maximise, but we require the same terms) so need expresion for x y and L

therfore I tried use similar triangles:

y / ((L-x)/2) = sqrt ((L^2)-(L/2)^2) / (L/2)

however this seems too complex and feel I have missed something much simpler...

I also tried using tan a = y/ ((L-x)/2)

however with tan a in the expression I ant maximise it as I dont know the angle.

The ANs is L/2 , sqrt(3)L/4

 

[skeeter]

hope you have a sketch drawn ...

base of the rectangle = x

what is left of the base of the equilateral triangle is L-x (the part that is not the base of the rectangle).

note that two small 30-60-90 triangles in both lower corners of the equilateral triangle will be formed with their "short" legs having length (L-x)/2

height of the rectangle = longer leg of the small 30-60-90 triangle = (L-x)/2*sqrt(3)

area of the rectangle will be ...

A(x) = sqrt(3)/2*x(L-x) = sqrt(3)/2*(xL - x²)

take it from here?

 

[ku1005]

cheers skeeter...


I TOTALLY forgot about the angles of the equilateral triangle being 60 -60 -60 and therefore resulting in common tiangles which i can use....simple prob now!!...i did it methods...thanks for ur help!!!