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...
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