Find the angle (in degrees) between two adjacent slant faces of a hexagonal pyramid which minimizes the lateral surface area of the pyramid when the volume is held constant.
Express the answer to two decimal places.
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Let O P be the height H of the hexagonal pyramid.
Let E : ( 2 − x , 2 3 , 0 ) , A : ( 2 − x , 2 − 3 , 0 ) , F : ( − x , 0 , 0 ) , P : ( 0 , 0 , H ) .
F P = x i + 0 j + H k , E P = 2 x i − 2 3 x j + H k , A P = 2 x i + 2 3 x j + H k
u = F P × E P = 2 3 x H i − 2 x H j − 2 3 x 2 k and v = F P × A P = 2 − 3 x H i − 2 x H j + 2 3 x 2 k
u ⋅ v = 4 − x 2 ( 2 H 2 + 3 x 2 ) and ∣ u ∣ = ∣ v ∣ = 2 x 4 H 2 + 3 x 2 ⟹ cos ( θ ) = ∣ u ∣ ∣ v ∣ u ⋅ v = 4 H 2 + 3 x 2 − ( 2 H 2 + 3 x 2 ) , where θ is the angle between the two adjacent slant faces.
The height of each of the six equilateral triangles of the hexagon is h ∗ = 2 3 x ⟹
A h e x a g o n = 2 3 3 x 2 ⟹ V p = 2 3 x 2 H
The slant height of the hexagonal pyramid s = h ∗ 2 + H 2 = 2 3 x 2 + 4 H 2 ⟹ the lateral surface S p = 3 x s = 2 3 x 3 x 2 + 4 H 2 .
V p = K ( c o n s t a n t ) ⟹ H = 3 x 2 2 K ⟹ S p ( x ) = 2 x 3 9 x 2 + 1 6 K 2 ⟹
d x d S p = 3 ( x 2 9 x 6 + 1 6 K 2 9 x 6 − 8 K 2 ) = 0 ⟹ x = ( 9 8 K 2 ) 6 1
Pick x = ( 2 6 K 2 ) 6 1 < ( 9 8 K 2 ) 6 1 ⟹ d x d S p < 0
Pick x = ( K 2 ) 6 1 > ( 9 8 K 2 ) 6 1 ⟹ d x d S p > 0 ⟹ x = d x d S p minimizes S p ( x ) .
Let u = ( 9 8 K 2 ) 6 1 ⟹ x = u 6 1 ⟹ H = 3 u 3 1 2 K
Using the above cos ( θ ) = 4 H 2 + 3 x 2 − ( 2 H 2 + 3 x 2 ) ⟹ cos ( θ ) = 1 6 K 2 + 9 u − ( 8 K 2 + 9 u ) = 3 − 2 ⟹ θ = 1 3 1 . 8 1 ∘