標題:
Question (Applications of Differentiation)
發問:
圖片參考：http://i33.tinypic.com/zladsw.jpgThe figure shows a rectangular picture of area A cm2 mounted on a rectangular piece of cardboard of area 3600 cm2 with sides of length x cm and y cm. The top, bottom and side margins are 12 cm, 13 cm and 8 cm wide respectively.(a) Find A in terms of x.(b) Show that... 顯示更多 圖片參考：http://i33.tinypic.com/zladsw.jpg The figure shows a rectangular picture of area A cm2 mounted on a rectangular piece of cardboard of area 3600 cm2 with sides of length x cm and y cm. The top, bottom and side margins are 12 cm, 13 cm and 8 cm wide respectively. (a) Find A in terms of x. (b) Show that the largest value of A is 1600. (c) (i) Find the range of values of x for which A decreases as x increases. (ii) If x ≦ 50, find the largest value of A. (d) If 4/9 < x/y < 9/16, find the range of values of x and the largest value of A.
a. Now, consider the cardboard: xy = 3600 y = 3600 / x Consider the picture: (y - 12 - 13)(x - 8 - 8) = A (y - 25)(x - 16) = A xy - 25x - 16y + 400 = A 3600 - 25x - 16(3600 / x) + 400 = A A = -25x + 4000 - 57 600/x b. dA/dx = -25 + 57 600/x2 d2A/dx2 = -115 200/x3 Set dA/dx = 0 -25 + 57 600/x2 = 0 x = 48 or -48 (rejected) (Since x > 0) When x = 48, d2A/dx2 < 0 So, max. of A is obtained when x = 48 So, the larger value of A = -25(48) + 4000 - 57 600/48 = 1600 c.i. As A attains a maximum when x = 48. And there is only one turning point. So, for x > 48, A decreases as x increases. ii. For x >= 50, A decreases as x increases. So, A is largest when x = 50 that is, max. of A = -25(50) + 4000 - 57 600/50 = 1598 d. 4/9 < x/y < 9/16 4y/9 < x < 9y/16 4/9 (3600/x) < x < 9/16 (3600/x) 1600/x < x < 2025/x 1600 < x2 < 2025 40 < x < 45 (Since x > 0) For this region, A increases as x increases. So, the largest value of A is obtained when x = 45 Largest value of A = -25(45) + 4000 - 57 600/45 = 1595
其他解答:
a) A = (x - 16)(y -25). But xy = 3600, thus A = (x - 16)(3600/x - 25) = 3600 - 25x - 57600/x + 400 = 4000 - 25x - 57600/x. b) dA/dx = -25 + 57600/x^2. Put it to zero, we get x^2 = 57600/25 = (240/5)^2 x = 240/5 = 48, which is a max. since d^2A/dx^2 < 0. So A max. = (48 - 16)(3600/48 - 25) = 32 x 50 = 1600. c) When A = (x -16)(y -25) = 0. x = 16 or y = 25 (that is x = 3600/25 = 144). So for A decreases as x increases, 48 < x < 144. d) 4/9 < x/y < 9/16 and xy = 3600, that is 4/9 < x^2/3600 < 9/16 1600< x^2 < 2025 40< x < 45 since x is positive. Largest value of A is when x = 45 because A increase as x increases. That is A = (45 - 16)(80 - 25) = 29 x 55 =1595.
Question (Applications of Differentiation)
發問:
圖片參考：http://i33.tinypic.com/zladsw.jpgThe figure shows a rectangular picture of area A cm2 mounted on a rectangular piece of cardboard of area 3600 cm2 with sides of length x cm and y cm. The top, bottom and side margins are 12 cm, 13 cm and 8 cm wide respectively.(a) Find A in terms of x.(b) Show that... 顯示更多 圖片參考：http://i33.tinypic.com/zladsw.jpg The figure shows a rectangular picture of area A cm2 mounted on a rectangular piece of cardboard of area 3600 cm2 with sides of length x cm and y cm. The top, bottom and side margins are 12 cm, 13 cm and 8 cm wide respectively. (a) Find A in terms of x. (b) Show that the largest value of A is 1600. (c) (i) Find the range of values of x for which A decreases as x increases. (ii) If x ≦ 50, find the largest value of A. (d) If 4/9 < x/y < 9/16, find the range of values of x and the largest value of A.
此文章來自奇摩知識+如有不便請留言告知
最佳解答:a. Now, consider the cardboard: xy = 3600 y = 3600 / x Consider the picture: (y - 12 - 13)(x - 8 - 8) = A (y - 25)(x - 16) = A xy - 25x - 16y + 400 = A 3600 - 25x - 16(3600 / x) + 400 = A A = -25x + 4000 - 57 600/x b. dA/dx = -25 + 57 600/x2 d2A/dx2 = -115 200/x3 Set dA/dx = 0 -25 + 57 600/x2 = 0 x = 48 or -48 (rejected) (Since x > 0) When x = 48, d2A/dx2 < 0 So, max. of A is obtained when x = 48 So, the larger value of A = -25(48) + 4000 - 57 600/48 = 1600 c.i. As A attains a maximum when x = 48. And there is only one turning point. So, for x > 48, A decreases as x increases. ii. For x >= 50, A decreases as x increases. So, A is largest when x = 50 that is, max. of A = -25(50) + 4000 - 57 600/50 = 1598 d. 4/9 < x/y < 9/16 4y/9 < x < 9y/16 4/9 (3600/x) < x < 9/16 (3600/x) 1600/x < x < 2025/x 1600 < x2 < 2025 40 < x < 45 (Since x > 0) For this region, A increases as x increases. So, the largest value of A is obtained when x = 45 Largest value of A = -25(45) + 4000 - 57 600/45 = 1595
其他解答:
a) A = (x - 16)(y -25). But xy = 3600, thus A = (x - 16)(3600/x - 25) = 3600 - 25x - 57600/x + 400 = 4000 - 25x - 57600/x. b) dA/dx = -25 + 57600/x^2. Put it to zero, we get x^2 = 57600/25 = (240/5)^2 x = 240/5 = 48, which is a max. since d^2A/dx^2 < 0. So A max. = (48 - 16)(3600/48 - 25) = 32 x 50 = 1600. c) When A = (x -16)(y -25) = 0. x = 16 or y = 25 (that is x = 3600/25 = 144). So for A decreases as x increases, 48 < x < 144. d) 4/9 < x/y < 9/16 and xy = 3600, that is 4/9 < x^2/3600 < 9/16 1600< x^2 < 2025 40< x < 45 since x is positive. Largest value of A is when x = 45 because A increase as x increases. That is A = (45 - 16)(80 - 25) = 29 x 55 =1595.
文章標籤
全站熱搜
留言列表
