A.math (maxima and minima) (20)

2007-05-04 5:45 am
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=)

回答 (1)

2007-05-09 10:15 am
✔ 最佳答案
#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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