The math help thread

Find the maximum volume of a right-circular cone placed upside-down in a right circular cone of radius R and height H. http://img451.imageshack.us/img451/6751/...ionfi3.jpg

Fun stuff this be.

Volume of a right-circular cone = 1/3*Π*r^2*h

The dimensions of the upside-down cone can be written in terms of the standing cone's dimensions.

Code:

r = r; 0 ≤ r ≤ R
h = H - H/R*r
  = H*(1 - r/R)

V = 1/3*Π*R^2*H; volume of the large cone
v = 1/3*Π*r^2*H*(1 - r/R); volume of the small cone

Then we find out at which value r within 0 ≤ r ≤ R can yield the largest coefficient value for the other two variables, H and R. To do this, we take the derivative of the volume function and find out where it equals 0. Points where the derivative equals 0 indicates minimums and maximums.

Code:

v' = 2/3*Π*r*H - Π*r^2*H/R
v' = Π*r*H*(2/3 - r/R)

If r = 2/3*R, [color=Red]v = 4/81*Π*H[/color]
If r = R, v = 0
If r = 0, v = 0

Find the maximum area of a rectangle that can be circumscribed about a given rectangle with length L and width W.
http://img685.imageshack.us/img685/530/rectyrec.png

Now this one is considerably trickier.

To determine the new dimensions of the circumscribed rectangle based on the original rectangle's dimensions L and W, and the variable θ, we'll apply geometry and trigonometry.

Code:

sin(θ) = o/h
cos(θ) = a/h
For both cases, h will either be L or W.

X = sin(θ)*W + cos(θ)*L; new rectangle's length
Y = sin(θ)*L + cos(θ)*W; new rectangle's width

V = X*Y
  = sin(θ)^2*L*W + cos(θ)^2*L*W + sin(θ)*cos(θ)*(W^2 + L^2)
  = L*W + sin(θ)*cos(θ)*(W^2 + L^2)

We'll do the same trick as with the cone problem: find the derivative of the volume function based off of θ.

Code:

V' = (-sin(θ)^2 + cos(θ)^2)*(W^2 + L^2)

If θ = Π/4 = 45°, V' = 0.

[color=Red]V = L*W + 1/2(W^2 + L^2)[/color]