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A.math (maxima and minima) (20)
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1) given that a area of a triangle is (bp^2)/(2p-2a) , find the least value of the triangle in terms of a and b. 2)V=(1/3).(pi).(y^3).(sinx)^2 . (cosx) y is constant, x varies a)find the value of x for V is maximum. b)the max. value of V in terms of x, y thz=) 更新: sorry...least area of triangle*
最佳解答:
#1) Do a, b, p has any meaning for the triangle? Are they all variables, or some being constants? #2) First, if y is zero, then V is always zero regardless of x. So (a) is all x, (b) is 0 Now assume y is non-zero. a) dV/dx =(1/3).(pi).(y^3).[2(sinx)(cosx)^2 - (sinx)^3] = 0. So (sinx)[2(cosx)^2 - (sinx)^2] = 0 That is, either sinx = 0, or 2(cosx)^2 - (sinx)^2=0 The first case is x = n*pi. However, this will make V = 0. Thus this must be a minimum. The second case can be rewritten as 2 - (tanx)^2 = 0 (divide both sides by (cosx)^2. Therefore, (tanx) = + or - sqrt(2) x = + or - 0.9553 This will be a local maximum because 0 which is inbetween the two choices, is a local minimum. Therefore, Subsituting x = + or - 0.9553 into V, V=(1/3).(pi).(y^3).0.3849 = 0.4031.(y^3)
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