Probem29+Sophia

Both of you have good reasoning...Jana I disagree with your derivative...I will keep working on it as well. I suddenly find that the derivative cannot be factored...

Problem 29 A right circular cylinder is inscribed in a sphere of radius r. Find the largest possible surface area of the cylinder.

I used x as the radius of the cylinder, because r is the radius of the sphere First, we find the equation we need to optimize, which is the surface area of a cylinder, A = 2pi x^2 + 2pi x h To put this in terms of one variable, because r is a constant, and it right cylinder, and it is inscribed in the sphere, we can look at a triangle from the center of the sphere (also the center of the cylinder). Its sides will be r, x, and 1/2 h. so: x^2 + (1/2h)^2 = r^2 or h^2 = 4r^2 - 4x^2 h= rt(4r^2 - 4x^2) h = 2 rt(r^2-x^2)

With h in terms of r, we can return to the equation that needs to be optimized A = 2pi x^2 + 2pi x (2 rt(r^2-x^2)) = 2pi x^2 + 4pi x (r^2-x^2)^1/2 A' = 4pi x+ 4pi x(-2x / 2 rt(r^2-x^2)) + 4 pi (r^2-x^2)^1/2 0 = 4pi (x + x (-2x/ 2 rt(r^2-x^2)) + (r^2-x^2)^1/2) 0 = 4pi (x - x^2/ rt(r^2-x^2) + rt(r^2-x^2)) 0 = (x(rt(r^2-x^2)))/rt(r^2-x^2) - x^2/rt(r^2-x^2) + (r^2-x^2)/rt(r^2-x^2) 0=(xrt(r^2-x^2) - x^2 + r^2 - x^2) / rt (r^2-x^2) 0 = x rt(r^2-x^2) - 2x^2 + r^2 (2x^2 - r^2)/x = rt(r^2-x^2) ((2x^2 - r^2)/x)^2 = r^2-x^2 4x^4-4x^2r^2+ r^4 = x^2r^2-x^4 4x^4 - 4x^2r^2 + r^4 - x^2r^2 + x^4 = 0 5x^4 - 5x^2r^2 + r^4 = 0



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Yes -- stick with it. I used "2h" as the height (so the pythagorean theorem would work out with lower numbers) so my equation is a bit different...but maybe try to do this on paper. It is tedious -- you will end up with a //quartic// (degree 4) equation -- but it can still be factored. Don't forget that when you find this value (which will be x in terms of r) you will need to plug your critical value into the surface area equation, and then compare this value with what happens at the end behavior. Also don't forget to wonder at how this works out so neatly!!! Jana, why don't you finish this up -- if you do it on paper, and can scan it in, that might be easier!