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/(y−1)  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 + y² ] =

= 2x ·  √ [ x²/4 + x²/(x−1)² ] =

= x · √ [ x⁴ − 2x³ + 5x² ] / (x−1)

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

f(x) =  (x⁶ − 2x⁵ + 5x⁴) / (x−1)²

f ' (x) = [ (6x⁵ − 10x⁴ + 20x³) · (x−1) − 2 · (x⁶ − 2x⁵ + 5x⁴) ] / (x−1)³

f ' (x) = 4x³ · (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

z³ + 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

u³ v³ = −8/27

    So,  u³  and  v³  are solutions to  

w² − 2w − 8/27 = 0

w = 1 ± √ (1+8/27) = 1 ± √ (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...