微分應用題

Let height of pyramid = h
Let the perpendicular distance from the vertex of the pyramid to x = k
so by Pythagoras thm, k^2 = h^2 + (x/2)^2.
Area of one side of the pyramid = kx/2
Area of 4 sides of the pyramid = 4(kx/2) = 2kx
Total internal surface area of silo = 2kx + x^2
= 2x sqrt ( h^2 + x^2/4) + x^2 = 32 ......................... (1)
Volume of silo, V = pi x^2h/3 .........................(2)
From (1),
sqrt ( 4x^2h^2 + x^4) + x^2 = 32
4x^2h^2 + x^4 = (32 - x^2)^2
4x^2h^2 + x^4 = 1024 - 64x^2 + x^4
4x^2h^2 = 1024 - 64x^2 ....................... (3)
Differentiate w.r.t. x,
8x^2h (dh/dx) + 8xh^2 = - 128x
x^2h (dh/dx) + xh^2 = - 16x
dh/dx = - (16x + xh^2)/x^2h..................(4)
From (2), differentiate w.r.t. x,
dV/dx = pi [ x^2(dh/dx) + 2xh]/3
Put dV/dx = 0
x^2 ( dh/dx) + 2xh = 0
Sub (4) into it,
- x^2 (16x + xh^2)/x^2h + 2xh = 0
- (16 + xh^2)/h + 2xh = 0
2xh^2 = 16 + xh^2
xh^2 = 16
h^2 = 16/x
h = sqrt (16/x)
that means volume is a max. when h = sqrt (16/x) and surface area kept at 32m^2.
Sub into (3),
64x = 1024 - 64x^2
x = 16 - x^2
x^2 + x - 16 = 0
x = 3.53 m or - 4.53 (rej.)
h = sqrt (16/3.53) = 2.13 m
so max. volume = 3.1416 x 3.53^2 x 2.13/3 = 27.8 m^2.



2014-11-17 11:45:49 補充：
Alternatively, you can get h = sqrt (256 - 16x^2)/x from (3), so volume , V = pi x sqrt (256 - x^2)/3. Put dV/dx = 0 will give you the value of x when volume is a max.