Question (Applications of Differentiation)
2017/07/04 03:39
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Question (Applications of Differentiation)
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)... 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.
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
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Question (Applications of Differentiation)
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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)... 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.
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
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