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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.
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