jueves, 12 de enero de 2017

1287. Optimized pyramid

    Pepe Chapuzas has proposed an optimization challenge.

    Dear Teacher:
    Calculate the base side and the height of a square pyramid circumscribed on a cube whose edge measures 1 m in order that its lateral area be minimal. 

    This challenge was solved by Tita Batalla...

    Dear Teacher:
    If  x  and  y  are the base side and the height of our pyramid, then  1  and  y−1  are the base side and the height of the little pyramid that is above the cube. The two pyramids are similar, so... 

x = y/(y1)  and  y = x/(x−1)

    The area we want to optimize can be calculated with Pythagoras' theorem:

Lateral area = 
= (base semiperimeter) · (slant height) =
= 2x · √ [ (x/2)2 + y] =
= 2x ·  √ [ x2/4 + x2/(x−1)] =
= x · √ [ x− 2x3 + 5x] / (x−1)

    It's easier to optimize the function (Lateral area)2:

f(x) =  (x− 2x5 + 5x4) / (x−1)2
f ' (x) = [ (6x− 10x4 + 20x3) · (x−1) − 2 · (x− 2x5 + 5x4) ] / (x−1)3
f ' (x) = 4x· (x3 − 3x2 + 5x − 5) / (x−1)3

    Since  x > 1 , I have to solve the equation  x3 − 3x2 + 5x − 5 = 0 ... Here Ruffini's rule isn't useful... I need another tool: Cardano's method... If  x = z + 1 then...

(z+1)3 − 3(z+1)2 + 5(z+1) − 5 = 0
z+ 3z2 + 3z + 1 − 3z2 − 6z − 3 + 5z + 5 − 5 = 0
z3 + 2z − 2 = 0
    If  z = u + v 
(u+v)3 + 2(u+v) − 2 = 0
u3 + v3 + 3uv(u+v) = 2 − 2(u+v)
u3 + v3 = 2
uv = −2/3
uv3 = −8/27
    So,  u3  and  v3  are solutions to  
w2 − 2w − 8/27 = 0
w = 1 ± √ (1+8/27) = ± √ (35/27)
u = 3√ [ 1 + √ (35/27) ]
v = 3√ [ 1 − √ (35/27) ]
x = 1 + u + v = 1 + 3√ [ 1 + √ (35/27) ] + 3√ [ 1 − √ (35/27) ]

    I go on with a calculator:
x = 1.771 m
y = 2.297 m


    The reader can make the necessary checks...

No hay comentarios:

Publicar un comentario en la entrada