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Math Prob
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最佳解答:
Volume of a cone = πr^2h/3 = 36 => h = 108/(πr^2) ... (1) The slant height L = √(h^2 + r^2) L^2 = r^2 + [108/(πr^2)]^2 L^2 = r^2 + (11664/π^2)r^(-4) ...(2) The circumference of the circle (cup mouth) = 2πr When open up the cup, the slant height becomes the radius The arc length = Lθ = 2πr => θ = 2πr/L Area of the paper A = (L^2)θ/2 = (L^2)(2πr/L)/2 = πrL ... (3) dA/dr = πL + πr(dL/dr) dA/dr = 0 => dL/dr = -L/r ... (4) Differentiate (2) wrt x, 2L(dL/dr) = 2r - (46656/π^2)r^(-5) Use (4), 2L(-L/r) = 2r - (46656/π^2)r^(-5) -2L^2 = 2r^2 - (46656/π^2)r^(-4) -L^2 = r^2 - (23328/π^2)r^(-4) Use (2), - r^2 - (11664/π^2)r^(-4) = r^2 - (23328/π^2)r^(-4) 2r^2 = (11664/π^2)r^(-4) r^6 = 5832/π^2 r = 2.897 => h = 4.097 => L = 5.017 => A = 45.66 r = 3 => h = 3.82 => L = 4.857 => A = 45.78 r = 2.8 => h = 4.385 => L = 5.203 => A = 45.76 Hence when r = 2.897 and h = 4.097, the paper consumed is a minimum 2009-11-23 22:08:18 補充： Area of paper = (πL^2)(θ/2π) θ is the angle of the sector in radian Complete circle = 2π radian Hence area of paper = (L^2)θ/2 And θ has been obtained in previous step to be 2πr/L okay? 2009-11-23 22:11:50 補充： See this site with diagram : http://www.mathsteacher.com.au/year10/ch14_measurement/18_cone/20cone.htm
其他解答:
What about i use s=pi*r*sqrt(r^2+h^2) to solve the problem. Is it duable? 2009-11-23 20:56:58 補充： Area of the paper A = (L^2)θ/2 = (L^2)(2πr/L)/2 = πrL ... I don't get this step.
Math Prob
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此文章來自奇摩知識+如有不便請留言告知
A cone-shaped paper drinking cup is to be made to hold 36 cm^3 of water. Find the height and radius of the cup that will use the smallest amount of paper.最佳解答:
Volume of a cone = πr^2h/3 = 36 => h = 108/(πr^2) ... (1) The slant height L = √(h^2 + r^2) L^2 = r^2 + [108/(πr^2)]^2 L^2 = r^2 + (11664/π^2)r^(-4) ...(2) The circumference of the circle (cup mouth) = 2πr When open up the cup, the slant height becomes the radius The arc length = Lθ = 2πr => θ = 2πr/L Area of the paper A = (L^2)θ/2 = (L^2)(2πr/L)/2 = πrL ... (3) dA/dr = πL + πr(dL/dr) dA/dr = 0 => dL/dr = -L/r ... (4) Differentiate (2) wrt x, 2L(dL/dr) = 2r - (46656/π^2)r^(-5) Use (4), 2L(-L/r) = 2r - (46656/π^2)r^(-5) -2L^2 = 2r^2 - (46656/π^2)r^(-4) -L^2 = r^2 - (23328/π^2)r^(-4) Use (2), - r^2 - (11664/π^2)r^(-4) = r^2 - (23328/π^2)r^(-4) 2r^2 = (11664/π^2)r^(-4) r^6 = 5832/π^2 r = 2.897 => h = 4.097 => L = 5.017 => A = 45.66 r = 3 => h = 3.82 => L = 4.857 => A = 45.78 r = 2.8 => h = 4.385 => L = 5.203 => A = 45.76 Hence when r = 2.897 and h = 4.097, the paper consumed is a minimum 2009-11-23 22:08:18 補充： Area of paper = (πL^2)(θ/2π) θ is the angle of the sector in radian Complete circle = 2π radian Hence area of paper = (L^2)θ/2 And θ has been obtained in previous step to be 2πr/L okay? 2009-11-23 22:11:50 補充： See this site with diagram : http://www.mathsteacher.com.au/year10/ch14_measurement/18_cone/20cone.htm
其他解答:
What about i use s=pi*r*sqrt(r^2+h^2) to solve the problem. Is it duable? 2009-11-23 20:56:58 補充： Area of the paper A = (L^2)θ/2 = (L^2)(2πr/L)/2 = πrL ... I don't get this step.
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