1) M = 2∏rh + ∏r^{2} V = 100 = ∏r^{2}h h = 100/∏r^{2} M = 200/r + ∏r^{2} M’ = 200/r^{2} + 2∏r = 0 = (2∏r^{3} – 200)/r^{2} ∏r^{3} = 100 r = ^{ 3}√(100/∏) C = 2∏r = ^{ 3}√(100/∏)(2∏)
2)
S = 2LW + 2WH + 2HL H = 5W S = 2LW + 10W^{2} + 10LW = 12LW + 10W^{2} (There are still too many variables) V = 1800 = LWH = 5LW^{2} L = 360/W^{2} S = 4320/W + 10W^{2} S’ = 4320/W^{2} + 20W = 0 = (20W^{3} – 4320)/W^{3} 20W^{3} = 4320 W = ^{ 3}√(216) = 6 H = 5(6) = 30 1800 = 6*30*L L = 10 3) D^{2} = (x  x_{1})^{2} + (y  y_{1})^{2} = (x)^{2} + (y + 4)^{2} y = x^{2}  59/2 D^{2} = x^{2} + (x^{2}  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. (D^{2})' = 2x + 4x^{3}  102x = 4x(x + 5)(x  5) Critical distances occur when x = 0, 5, 5. (D^{2})'' = 12x^{2}  100 x = 0, (D^{2})'' is negative, distance at a local max. x = 5, 5, (D^{2})'' 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.
