Minimizing Surface Area

Calculus Level 5

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.


The answer is 131.81.

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1 solution

Rocco Dalto
Dec 16, 2017

Let O P OP be the height H H of the hexagonal pyramid.

Let E : ( x 2 , 3 2 , 0 ) E: (\dfrac{-x}{2}, \dfrac{\sqrt{3}}{2}, 0) , A : ( x 2 , 3 2 , 0 ) A: (\dfrac{-x}{2}, \dfrac{-\sqrt{3}}{2}, 0) , F : ( x , 0 , 0 ) F: (-x,0,0) , P : ( 0 , 0 , H ) P: (0,0,H) .

F P = x i + 0 j + H k , E P = x 2 i 3 2 x j + H k , A P = x 2 i + 3 2 x j + H k \vec{FP} = x\vec{i} + 0\vec{j} + H\vec{k} , \:\ \vec{EP} = \dfrac{x}{2}\vec{i} - \dfrac{\sqrt{3}}{2} x\vec{j} + H\vec{k}, \:\ \vec{AP} = \dfrac{x}{2}\vec{i} + \dfrac{\sqrt{3}}{2} x\vec{j} + H\vec{k}

u = F P × E P = 3 2 x H i x H 2 j 3 2 x 2 k \vec{u} = \vec{FP} \times \vec{EP} = \dfrac{\sqrt{3}}{2} xH \vec{i} -\dfrac{xH}{2}\vec{j} - \dfrac{\sqrt{3}}{2} x^2\vec{k} and v = F P × A P = 3 2 x H i x H 2 j + 3 2 x 2 k \vec{v} = \vec{FP} \times \vec{AP} = \dfrac{-\sqrt{3}}{2} xH \vec{i} -\dfrac{xH}{2}\vec{j} + \dfrac{\sqrt{3}}{2} x^2\vec{k}

u v = x 2 4 ( 2 H 2 + 3 x 2 ) \vec{u} \cdot \vec{v} = \dfrac{-x^2}{4} (2 H^2 + 3 x^2) and u = v = x 2 4 H 2 + 3 x 2 cos ( θ ) = u v u v = ( 2 H 2 + 3 x 2 ) 4 H 2 + 3 x 2 |\vec{u}| = |\vec{v}| = \dfrac{x}{2} \sqrt{4H^2 + 3x^2} \implies \cos(\theta) = \dfrac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \dfrac{-(2H^2 + 3x^2)}{4H^2 + 3x^2} , where θ \theta is the angle between the two adjacent slant faces.

The height of each of the six equilateral triangles of the hexagon is h = 3 2 x h^{*} = \dfrac{\sqrt{3}}{2} x \implies
A h e x a g o n = 3 3 2 x 2 V p = 3 2 x 2 H A_{hexagon} = \dfrac{3\sqrt{3}}{2} x^2 \implies V_{p} = \dfrac{\sqrt{3}}{2} x^2 H

The slant height of the hexagonal pyramid s = h 2 + H 2 = 3 x 2 + 4 H 2 2 s = \sqrt{{h^{*}}^2 + H^2} = \dfrac{\sqrt{3x^2 + 4H^2}}{2} \implies the lateral surface S p = 3 x s = 3 x 2 3 x 2 + 4 H 2 . S_{p} = 3xs = \dfrac{3x}{2} \sqrt{3x^2 + 4H^2}.

V p = K ( c o n s t a n t ) H = 2 K 3 x 2 S p ( x ) = 3 2 x 9 x 2 + 16 K 2 V_{p} = K(constant) \implies H = \dfrac{2K}{\sqrt{3} x^2} \implies S_{p}(x) = \dfrac{\sqrt{3}}{2x} \sqrt{9x^2 + 16K^2} \implies

d S p d x = 3 ( 9 x 6 8 K 2 x 2 9 x 6 + 16 K 2 ) = 0 \dfrac{dS_{p}}{dx} = \sqrt{3} (\dfrac{9x^6 - 8K^2}{x^2\sqrt{9x^6 + 16K^2}}) = 0 \implies x = ( 8 K 2 9 ) 1 6 x = (\dfrac{8 K^2}{9})^\dfrac{1}{6}

Pick x = ( K 2 2 6 ) 1 6 < ( 8 K 2 9 ) 1 6 d S p d x < 0 x = (\dfrac{K^2}{2^6})^\dfrac{1}{6} < (\dfrac{8 K^2}{9})^\dfrac{1}{6} \implies \dfrac{dS_{p}}{dx} < 0

Pick x = ( K 2 ) 1 6 > ( 8 K 2 9 ) 1 6 d S p d x > 0 x = d S p d x x = (K^2)^\dfrac{1}{6} > (\dfrac{8 K^2}{9})^\dfrac{1}{6} \implies \dfrac{dS_{p}}{dx} > 0 \implies x = \dfrac{dS_{p}}{dx} minimizes S p ( x ) . S_{p}(x).

Let u = ( 8 K 2 9 ) 1 6 x = u 1 6 H = 2 K 3 u 1 3 u = (\dfrac{8 K^2}{9})^\dfrac{1}{6} \implies x = u^\dfrac{1}{6} \implies H = \dfrac{2K}{\sqrt{3}u^\dfrac{1}{3}}

Using the above cos ( θ ) = ( 2 H 2 + 3 x 2 ) 4 H 2 + 3 x 2 \cos(\theta) = \dfrac{-(2H^2 + 3x^2)}{4H^2 + 3x^2} \implies cos ( θ ) = ( 8 K 2 + 9 u ) 16 K 2 + 9 u = 2 3 θ = 131.8 1 \cos(\theta) = \dfrac{-(8K^2 + 9u)}{16K^2+ 9u} = \dfrac{-2}{3} \implies \theta = \boxed{131.81^\circ}

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