<p>Maximize the are of the isosceles triangle inscribed in the circle of radius 6.</p>
<p>Picture</p>
<p>How the heck do you solve this?</p>
<p>The area is max when the triangle is [url=<a href="http://www.bloom-enterprises.com/Math/oproblem2.html%5Dequilateral%5B/url">http://www.bloom-enterprises.com/Math/oproblem2.html]equilateral[/url</a>]. </p>
<p>Basically, follow the work on that page until they prove that x=(sqrt(3))*a/2 (the critical number for the area equation). Then plug x and a (which is 6 in your case) into the orignial area equation to find max area.</p>
<p>Assuming that the equilateral is the maximized area I did the math and got 27<em>sqrt(3) seeing that the 2 corner angles are equal, and knowing that on an equilateral that total angle=60 you know that each sub angle must be 30. So then you see the 6 on the hypotenuse and know that the adjacent side (1/2 of the overall base) = 3</em>sqrt(3). Since that is half the base you discover that each side is therefore 6<em>sqrt(3), and use the pythagorean theorem again to find the height. (3</em>sqrt(3))^2 + h^2 = (6<em>sqrt(3))^2. Then you get h=9, and half base</em>height gives you the area. (3<em>sqrt(3))</em>9= 27*sqrt(3).</p>
<p>So actually as long as you knew that equilateral triangles have the max area... you dont really need derivatives.</p>
<p>I'm pretty sure that's right... can anyone check me?</p>
<p>yeah that's the answer my calculus teacher said we were supposed to get, except we had to work the problem out using derivatives, as tanman showed</p>
<p>tanman es correcto.</p>
<p>u guys have harder calc problems than i do... and i'm in bc, and i have a 95 average... i think i'm gonna get a 3 :(</p>