有a.maths 唔識做_3－ppi93gk88d的部落格

標題:

有a.maths 唔識做

發問:

Find the maximum and minimum values of siny - 2cos y + 3

最佳解答:

有 2 個辦法 Differentiation 同 Supp. Angel 1. diff Let g = sin y – 2cos y +3 dg / dy = cos y + 2 sin y d2g / dy2 = -sin y + 2 cos y put dg / dy = 0 and we have tan y = -1/2 that is we got 2 pair of answers (cos y, sin y) = (2/sqrt5, -1/sqrt5) or (-2/sqrt5, 1/sqrt5) Check with d2g / dy2, we have (cos y, sin y) = (2/sqrt5, -1/sqrt5) is the min pt and (cos y, sin y) = (-2/sqrt5, 1/sqrt5) is the max pt so the required value = 1/sqrt5 – 2* (-2/sqrt5) + 3 = 3 + sqrt5 2. supp. angel Let sin y – 2cos y = r cos (y + z), where z is a constant r^2 = 1^2 + 2^2 that is r = sqrt 5 [ 原本可以計埋 z 出黎, 不過題目無要求就算喇 ] sin y – 2cos y +3 = sqrt5 * cos (y + z) + 3 咁 max value 當然係cos (y + z) = 1 既時候亦即 3 + sqrt 5

其他解答:

Let f(y)=siny- 2cos y + 3 f'(y)=cosy+2siny f''(y)=-siny+2cosy Let f'(y)=0 cosy+2siny=0 tany=-1/2 we got 2 pair of answers (cos y, sin y) = (2/sqrt5, -1/sqrt5) or (-2/sqrt5, 1/sqrt5) 2008-05-12 20:55:45 補充： Check with f''(y), we have (cos y, sin y) = (2/sqrt5, -1/sqrt5) is the minimum point and (cos y, sin y) = (-2/sqrt5, 1/sqrt5) is the maximum point so the maximum value = 1/sqrt5 – 2* (-2/sqrt5) + 3 = 3 + sqrt5 minimum value=-1/sqrt5 – 2* (2/sqrt5) + 3 = 3 - 5/sqrt5