標題:
minimum surface of a can
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發問:
1. Assuminga fixed volume can of 785cm3, set up the equation of surface area Averses radius R.2. Atwhat radius is the surface area minimum? 3. Whatheight corresponds to the minimum surface area?4. Whatis the advantage of producing a can with minimum surface area? Thank you !!
最佳解答:
1. Let h be the height of the can and * is a multiplication sign Volume of cylinder = (pi) * R^2 * (h) where R = radius of the can 785= (pi)R^2 * (h) h = 785/(pi * R^2) Surface area of cylinder = circular areas of top and bottom + area of curved surface A = 2pi*(R^2) + 2pi*R*h A = 2pi*(R^2) + 2pi*R*[785/(pi * R^2)] A = 2pi*R^2 + 1570/R 2. Differentiate Area A with respect to R A = 2pi*R^2 + 1570*R^(-1) dA/dR = 4pi*R + 1570*(-1)*R^(-1-1) dA/dR = 4pi*R – 1570R^(-2) dA/dR = 4pi*R – 1570/R^2 Let dA/dR = 0 0 = 4pi*R – 1570/R^2 Multiply both sides by R^2 (0)*(R^2) = 4pi*R*( R^2)– 1570*R^2/R^2 0 = 4pi*R^3– 1570 4pi*R^3 = 1570 R^3 = 1570/4pi = 124.9366 R = 124.9366^(1/3) = 4.999 cm R = 5 cm You have to show that A is a minimum value when R = 5 cm Take second derivative d/dR(dA/dR) = d/dR[4pi*R – 1570R^(-2)] d2A/dR2 = 4pi + 4710/Rr^3 Since R is positive, so d2A/dR2 is always positive, hence area is a minimum value. Surface area is at minimum when radius of the can is 5 cm 3. h = 785/(pi * R^2) h = 785/(pi * 5^2) h = 9.99 cm h = 10 cm 4. Material used to produce the can will be at minimum since the area is minimum. Consequently, the cost of material will be at minimum. If the can is coated with a paint, so does the cost of the paint at minimum.
其他解答:
1. Let h be the height of can. (pi)R^2h = 785 h = 785/piR^2 A = 2piR^2 + 2piRh = 2piR^2 + 1570/R 2. dA/dR = 4piR - 1570/R^2 dA/dR = 0 4piR^3 - 1570 = 0 R = cuberoot(785/2pi) dA/dR R 0 R > cuberoot(785/2pi) So A attains minimum when R = cuberoot(785/2pi) cm 3. h = 785/[pi(785/2pi)^2/3] = cuberoot(3140/pi) cm 4. Cost of painting can reduce to minimum.
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