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As follows~ 圖片參考:http://i248.photobucket.com/albums/gg191/ncy_0916/amaths50.jpg?t=1208878573 圖片參考:http://i248.photobucket.com/albums/gg191/ncy_0916/amaths52.jpg?t=1208878698 2008-04-23 18:57:10 補充: 2)Let the area of the rectangle be A. A = (2a)(-a^2 + 12) = -2a^3 +24a dA/da = -6a^2 + 24 d^2 A / da^2 = -12a When dA/da = 0, 6a^2 = 24 a = 2 or -2(rejected) (d^2A / da^2) | a=2 = -24 < 0 So A is maximum when a = 2, the dimensions of the rectangle are 4×8
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1. A restangular container with a spuare base is made to contain 720m^3.The cost of materials for the base, top and four sides is $12/m^2 , $8/m^2 and 3/m^2 respectively. Find the dimensions of the container such that the total cost of materials is a minimum.2. A rectangle with its base on the x-asis is... 顯示更多 1. A restangular container with a spuare base is made to contain 720m^3. The cost of materials for the base, top and four sides is $12/m^2 , $8/m^2 and 3/m^2 respectively. Find the dimensions of the container such that the total cost of materials is a minimum. 2. A rectangle with its base on the x-asis is inscribed in a quadratic curve y= -x^2+12. Find the dimensions of the rectangle if its area is a maximum. 3. Four squares are cut from each corner of a rectangular metal sheet of legth 16cm and width 6cm. the remaining part is folded into an open box. Find the leugth of the squares being cut out so that the box has the greatest volume. 4. A right circular cone has a volume of 18兀cm^3. Let l and h be the slant height and the height of the cone respectively. A) Show that l^2=h^2+54/h. B) Hence, find the minimum value of l as h varies. 5. A window ABCDE of area 8m^2. ABCD is a rectangle of length 2x m and height hm and ADE is a semicircle. A) Expree h in terms of x. B) Find the value of x of the perimeter of the window is a minimum. 唔該幫幫手~最好全部做哂啦^^ 但係識幾題都ok最佳解答:
As follows~ 圖片參考:http://i248.photobucket.com/albums/gg191/ncy_0916/amaths50.jpg?t=1208878573 圖片參考:http://i248.photobucket.com/albums/gg191/ncy_0916/amaths52.jpg?t=1208878698 2008-04-23 18:57:10 補充: 2)Let the area of the rectangle be A. A = (2a)(-a^2 + 12) = -2a^3 +24a dA/da = -6a^2 + 24 d^2 A / da^2 = -12a When dA/da = 0, 6a^2 = 24 a = 2 or -2(rejected) (d^2A / da^2) | a=2 = -24 < 0 So A is maximum when a = 2, the dimensions of the rectangle are 4×8
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