Finding the Volume 3

Calculus Level 5

For each positive integer n n let b n b_{n} and c n c_{n} be real constants. Minimize the volume V n V_{n} of the region bounded between y = x 4 n + b n x 2 n + c n y = x^{4n} + b_{n}x^{2n} + c_{n} , x = 0 x= 0 and x = 1 x=1 , when it is revolved about the x x -axis.

If this volume V n V_{n} can be expressed as V n = ( β α λ n + β ) V_{n} = (\dfrac{\beta}{\alpha^{\lambda}n + \beta}) ( α λ n α ( α ∗ λ n + b ) ( α α n + β ) ) α ∗ π (\dfrac{\alpha^{\lambda} n^{\alpha}}{(\alpha * \lambda n + b)(\alpha^{\alpha} n + \beta)})^{\alpha} * \pi , where α , β \alpha,\beta and λ \lambda , are coprime positive integers, submit your answer as α + β + λ \alpha + \beta + \lambda .


The answer is 6.

Rocco Dalto by Rocco Dalto , 67, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rocco Dalto
Apr 23, 2018

The volume V n ( a , b ) = π ∫ 0 1 ( x 4 n + b x 2 n + c ) 2 d x = π ∫ 0 1 ( x 8 n + 2 b x 6 n + ( b 2 + 2 c ) x 4 n + 2 b c x 2 n + c 2 ) d x V_{n}(a,b) = \pi\int_{0}^{1} (x^{4n} + bx^{2n} + c)^2 dx = \pi\int_{0}^{1} (x^{8n} + 2bx^{6n} + (b^2 + 2c)x^{4n} + 2bc x^{2n} + c^2)dx

= π ( x 8 n + 1 8 n + 1 + 2 b 6 n + 1 x 6 n + 1 + b 2 + 2 c 4 n + 1 x 4 n + 1 + 2 b c 2 n + 1 x 2 n + 1 + c 2 x ) 0 1 = = \pi(\dfrac{x^{8n + 1}}{8n + 1} + \dfrac{2b}{6n + 1}x^{6n + 1} + \dfrac{b^2 + 2c}{4n + 1}x^{4n + 1} + \dfrac{2bc}{2n + 1}x^{2n + 1} + c^2x)_{0}^{1} = π ( 1 8 n + 1 + 2 b 6 n + 1 + b 2 + 2 c 4 n + 1 + 2 b c 2 n + 1 + c 2 ) \pi(\dfrac{1}{8n + 1} + \dfrac{2b}{6n + 1} + \dfrac{b^2 + 2c}{4n + 1} + \dfrac{2bc}{2n + 1} + c^2)

⟹ ∂ V ∂ b = 2 π ( 1 6 n + 1 + b 4 n + 1 + c 2 n + 1 ) = 0 \implies \dfrac{\partial{V}}{\partial{b}} = 2\pi(\dfrac{1}{6n + 1} + \dfrac{b}{4n + 1} + \dfrac{c}{2n + 1}) = 0 and ∂ V ∂ c = 2 π ( 1 4 n + 1 + b 2 n + 1 + c ) = 0 \dfrac{\partial{V}}{\partial{c}} = 2\pi(\dfrac{1}{4n + 1} + \dfrac{b}{2n + 1} + c) = 0

⟹ 1 4 n + 1 b + 1 2 n + 1 c = − 1 6 n + 1 \implies \dfrac{1}{4n + 1}b + \dfrac{1}{2n + 1}c = \dfrac{-1}{6n + 1} and 1 2 n + 1 b + c = − 1 4 n + 1 \dfrac{1}{2n + 1}b + c = \dfrac{-1}{4n + 1}

Solving the system we obtain b = − 2 ( 2 n + 1 ) 6 n + 1 b = \dfrac{-2(2n + 1)}{6n + 1} and c = 2 n + 1 ( 4 n + 1 ) ( 6 n + 1 ) c = \dfrac{2n + 1}{(4n + 1)(6n + 1)} .

∂ 2 V ∂ b 2 = 2 π 4 n + 1 > 0 \dfrac{\partial^2{V}}{\partial{b^2}} = \dfrac{2\pi}{4n + 1} > 0 and ∂ 2 V ∂ c 2 = 2 π > 0 \dfrac{\partial^2{V}}{\partial{c^2}} = 2\pi > 0 and ∂ ∂ b ( ∂ V ∂ c ) = ∂ ∂ c ( ∂ V ∂ b ) = 2 π 2 n + 1 \dfrac{\partial}{\partial{b}}(\dfrac{\partial{V}}{\partial{c}}) = \dfrac{\partial}{\partial{c}}(\dfrac{\partial{V}}{\partial{b}}) = \dfrac{2\pi}{2n + 1}

The hessian matrix M = ∣ 2 π 4 n + 1 2 π 2 n + 1 2 π 2 n + 1 2 π ∣ M= \begin{vmatrix}{\dfrac{2\pi}{4n + 1}} && {\dfrac{2\pi}{2n + 1}} \\ {\dfrac{2\pi}{2n + 1}} && {2\pi}\end{vmatrix} and det ⁡ ( M ) = 16 n 2 π 2 ( 4 n + 1 ) ( 2 n + 1 ) 2 > 0 ⟹ \det(M) = \dfrac{16n^2\pi^2}{(4n + 1)(2n + 1)^2} > 0 \implies we have a minimum at ( − 2 ( 2 n + 1 ) 6 n + 1 , 2 n + 1 ( 4 n + 1 ) ( 6 n + 1 ) (\dfrac{-2(2n + 1)}{6n + 1},\dfrac{2n + 1}{(4n + 1)(6n + 1})

After simplifying we obtain: V n = 64 n 4 π ( 8 n + 1 ) ( 6 n + 1 ) 2 ( 4 n + 1 ) 2 = ( 1 2 3 n + 1 ) ( 2 3 n 2 ( 2 ∗ 3 n + 1 ) ( 2 2 n + 1 ) ) 2 ∗ π = V_{n} = \dfrac{64n^4\pi}{(8n + 1)(6n + 1)^2(4n + 1)^2} = (\dfrac{1}{2^3 n + 1}) (\dfrac{2^3 n^2}{( 2* 3 n +1)(2^2 n + 1)})^2 * \pi = ( β α λ + β ) ( α λ n α ( α ∗ λ n + b ) ( α α n + β ) ) α ∗ π ⟹ α + β + λ = 6 (\dfrac{\beta}{\alpha^{\lambda} + \beta}) (\dfrac{\alpha^{\lambda} n^{\alpha}}{(\alpha * \lambda n + b)(\alpha^{\alpha} n +\beta)})^{\alpha} * \pi \implies \alpha + \beta + \lambda = \boxed{6}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...