Answers Optimization

1)
M = 2∏rh + ∏r2
V = 100 = ∏r2h     h = 100/∏r2
M = 200/r + ∏r2
M = -200/r2 + 2∏r = 0 = (2∏r3 200)/r2
∏r3 = 100    r = 3√(100/∏)
C = 2∏r = 3√(100/∏)(2∏)
2)
S = 2LW + 2WH + 2HL
H = 5W
S = 2LW + 10W2 + 10LW = 12LW + 10W2
(There are still too many variables)
V = 1800 = LWH = 5LW2      L = 360/W2
S = 4320/W + 10W2
S = -4320/W2 + 20W = 0 = (20W3 4320)/W3
20W3 = 4320    W = 3√(216) = 6
H = 5(6) = 30
1800 = 6*30*L
L = 10

3)
D2 = (x - x1)2 + (y - y1)2 = (x)2 + (y + 4)2
y = x2 - 59/2 
D2 = x2 + (x2 - 51/2)2
At this point, since we are going to set the derivative to 0, we don't need to square root the equation. 0 is the same as the square root of 0. 
(D2)' = 2x + 4x3 - 102x = 4x(x + 5)(x - 5)
Critical distances occur when x = 0, 5, -5. 
(D2)'' = 12x2 - 100
x = 0, (D2)'' is negative, distance at a local max. 
x = 5, -5, (D2)'' is positive, distance at a local min.
Minimums at x = 5, -5.
Refer to Critical Points if you did not understand this second
derivative test.

 

Copyright 2001 Bruce
Email: