Probem25Alex+G,+Elie

Question: Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r.

To do this I think the key is to first put the whole diagram on a set of axes. Then, the only issue is to find the height and base of the triangle in terms of r. With this, one can use the formula A=.5BH

Any ideas?

I got that the base equals 2y and the height is r+x.  Elie: Using a triangle where the base equals 2y and the height equals r+x, the Objective Equation that we're maximizing is: A=bh/2 A=[2y(r+x)]2 A=y(r+x)



Second Equation using pythagorean theorem involves triangle made from drawing radius from the base-corner of the isosceles triangle. New triange has sides of length x,y, and r. The equation is, therefore, x^2 + y^2=r^2. Simplifying, x=squareroot(r^2 -y^2)

Substituting into objective equation....

A=y(r+sqt(r^2-y^2) A=yr+ysqrt(r^2-y^2)

Find critical values by taking 1st deriv. and setting it equal to zero. It gets a little messy here, just warning you.

A'=r+ sqrt(r^2-y^2)-[y^2/(sqrt(r^2-y^2)]=0 r(sqrt(r^2-y^2)+r^2-y^2-y^2=0 r(sqrt(r^2-y^2)+r^2-2y^2=0 sqrt(r^2-y^2)+r-(2y^2)/r=0 sqrt(r^2-y^2)=-r-(2y^2)/r r^2-y^2=r^2-4y^2+(4y^4)/r^2 -y^2+4y^2-(4y^4/r^2)=0 3y^2=4y^4/r^2 3r^2=4y^2 y=sqrt(3r^2/4) y=3r/sqrt2 :I actually get (√3 / 2) r for y...so the base is r √3, the height ends up being 1.5r. I also could be wrong!! (ms sweeney)

A(3r/sqrt2)=3r/sqrt2v(r)+[3r/sqrt2(r^2-3r/sqrt2)] =3r^2/sqrt2+[3r/sqrt2(r^2-3r/sqrt2)]

Now, we need to test the endpoints which are zero and 2r. 2r because the triange is inscribed in the circle so its sides must be shorter than the diameter.

lim A= 0 y->0

lim A= 2r^2+ysqrt(-3r^2) this is impossible since there will be a negative under the square root sign. y->2r

Therefore y=3r/sqrt2 must be the correct value for y.

The dimensions of the triangle would be, as a result, base=6r/sqrt(-3r^2) and to find the height use the second equation

x=squareroot(r^2 -y^2) x=sqrt(r^2-9r^2/2) height= r-sqrt(r^2-9r^2/2)

I know this is ugly and different than the book's answer, but I couldn't find a better solution.....sorry.