m28+Danny+G,+Marlee

Very good!

[|Diagram]

We are trying to maximize the equation for the volume of the right circular cylinder V= pi(r^2)h, that is inscribed in a cone over the interval r = [0,R]

The first step is to find an equation that relates the height and radius of the cylinder to the height and radius of the inscribed cylinder. To do this, we can show that there are similar triangles, and therefore (H-h)/r = H/R

(H-h)/r = H/R -h+H = rH/R h = -rH/R + H

Now we can use this equation to plug into the original function

V = pi(r^2)(-rH/R +H) V = pi(-r^3H/R + r^2H)

Now we must take the derivative and set it equal to zero in order to find critical points and local extrema

V' = pi [-3r^2(H/R) + 2rH] 0 = pi(rH)[-3r/R + 2] 0 = -3r/R + 2 -2 = -3r/R -2R = -3r r =(2/3)R

Now we must see if this is a local extrema, by testing the activity of the endpoints within the interval V(0) = pir^2(-rH/R + H) V(0) = pi(0^2)(OH/R + H) V(0) = 0

V(R) = pi(R^2)(-RH/R + H) V(R) = pi (-R^2H + R^2H) V(R) = pi (0) V(R) = 0

V(2R/3) = pi [(-2R/3)^3H/R + (2R/3)^2H] V(2R/3) = pi [(-8/27)R^3H/R + (4/9)R^2H) V(2R/3) = pi [(-8/27)R^2H + (4/9)R^2H] V(2R/3) = pi (4/27)R^2H

Therefore, we see that in order to have the cylinder maximize its area within the cone, the radius of the cone must be (2/3)R, or (2/3) of the original radius of the cone.