Calculus Optimization Problem - Need Serious Help

[Del Murder]

Reading math on a forum is murder. I worked it out at work today and got what everone else got, even though I still couldn't prove why that angle is a right angle. Anyway, it is, but that wasn't necessary for your method, which is good.

Glad you figured it out on your own, though. Good work!

[crono_logical]

Since when did equilateral triangles have right angles in?

[escobert]

Never. Yay the only math I was good at was geometry, I should have stopped after geometry. Damn PSU making me pass algerbra II before i could be accepted

[Del Murder]

Since when did equilateral triangles have right angles in?

Since never. Maybe you should look at my drawing and see what I referred to before making a smart alec comment, Archie!

[Auronhart]

Hmm, too bad I missed out, I like math problems.

[SeeDRankLou]

even though I still couldn't prove why that angle is a right angle.

In your picture, we know that the (b/2)(8-h)z triangle is a right triangle. And with the problem solved we know that b/2=2√3 and that 8-h=2, therefore z=4. Also with the problem solve one can also derive that c=4√3. So (d=diameter), if the zcd triangle is a right triangle, then the Pythagorean theorem should work. And behold 4² + (4√3)² = 8². Therefore, zcd is a right triangle. Without having solved the problem I'm sure you can do what Kirobaito did and put everything in terms of h and use the same logic as above.

[Endless]

Any triangle whose vertex is on the edge of the circle and has the diameter as the base is a right triangle. You can consider it a given.

Graphically, it's piss-easy to demonstrate it. Take your vertex, draw the line that starts from it, crosses the center of the disk, and mark a vertex where it touches the opposite edge. You now have 4 vertices, the original one, the one you just drew, and the two on the diameter. any quadrilatere with its diagonals of equal lengths, cutting each others in their middle is a rectangle, which means, right angles.