Minimizing Resultant Volume

Calculus Level pending

Let a a , b b , and c c be real constants. Minimize the volume of the region bounded between y = x 3 + a x 2 + b x + c y = x^3 + ax^2 +bx + c , x = 0 x= 0 and x = 1 x=1 , when it is revolved about the x x -axis.

If this volume can be expressed as m n π \dfrac mn \pi , where m m and n n are coprime positive integers, submit your answer as m + n m+n .


The answer is 2801.

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1 solution

Rocco Dalto
Nov 10, 2016

The volume V ( a , b , c ) = π 0 1 ( x 3 + a x 2 + b x + c ) 2 d x = {\bf V(a,b,c) = \pi * \int_{0}^{1} (x^3 + ax^2 + bx + c)^2 dx = }

π 0 1 ( x 6 + 2 a x 5 + ( a 2 + 2 b ) x 4 + 2 ( a b + c ) x 3 + ( 2 a c + b 2 ) x 2 + 2 b c x + c 2 ) d x = {\bf \pi * \int_{0}^{1} (x^6 + 2a x^5 + (a^2 + 2b)x^4 + 2(ab + c)x^3 + (2ac + b^2)x^2 + 2bcx + c^2) \: dx = }

π ( x 7 7 + a 3 x 6 + a 2 + 2 b 5 x 5 + {\bf \pi * (\frac{x^7}{7} + \frac{a}{3} x^6 + \frac{a^2 + 2b}{5} x^5 + } a b + c 2 x 4 + 2 a c + b 2 3 x 3 + b c x 2 + c 2 x ) 0 1 = {\bf \frac{ab + c}{2} x^4 + \frac{2ac + b^2}{3} x^3 + bc x^2 + c^2 x)|_{0}^{1} = }

π ( 1 7 + a 3 + a 2 + 2 b 5 + a b + c 2 + 2 a c + b 2 3 + b c + c 2 ) {\bf \pi * (\frac{1}{7} + \frac{a}{3} + \frac{a^2 + 2b}{5} + \frac{ab + c}{2} + \frac{2ac + b^2}{3} + bc + c^2) }

V a = 1 3 + 2 a 5 + b 2 + 2 c 3 = 0 {\bf \frac{\partial V}{\partial a} = \frac{1}{3} + \frac{2a}{5} + \frac{b}{2} + \frac{2c}{3} = 0 }

V b = 2 5 + a 2 + 2 b 3 + c = 0 {\bf \frac{\partial V}{\partial b} = \frac{2}{5} + \frac{a}{2} + \frac{2b}{3} + c = 0 }

V c = 1 2 + 2 a 3 + b + 2 c = 0 {\bf \frac{\partial V}{\partial c} = \frac{1}{2} + \frac{2a}{3} + b + 2c = 0 }

{\bf \implies }

12 a + 15 b + 20 c = 10 {\bf 12a + 15b +20c = -10 }

15 a + 20 b + 30 c = 12 {\bf 15a + 20b + 30c = -12 }

4 a + 6 b + 12 c = 3 {\bf 4a + 6b + 12 c = -3 }

{\bf \implies }

3 b + 16 c = 1 {\bf 3b + 16c = 1 }

10 b + 60 c = 3 {\bf 10b + 60c = 3 }

{\bf \implies }

c = 1 20 , b = 3 5 a = 3 2 {\bf c = \frac{-1}{20}, b = \frac{3}{5} \implies a = \frac{-3}{2} }

and,

2 V a 2 = 2 5 > 0 {\bf \frac{\partial^2 V}{\partial a^2} = \frac{2}{5} > 0 }

2 V b 2 = 2 3 > 0 {\bf \frac{\partial^2 V}{\partial b^2} = \frac{2}{3} > 0 }

2 V c 2 = 2 > 0 {\bf \frac{\partial^2 V}{\partial c^2} = 2 > 0 }

and,

the Hessian matrix M = [ 2 5 0 0 0 2 3 0 0 0 2 ] {\bf M = [ \begin{array}{ccc} \frac{2}{5} & 0 & 0 \\ 0 & \frac{2}{3} & 0 \\ 0 & 0 & 2 \\ \end{array} ] }

d e t ( M ) > 0 {\bf det(M) > 0 \implies }

we have a minimum value at ( 3 2 , 3 5 , 1 20 ) {\bf (\frac{-3}{2}, \frac{3}{5}, \frac{-1}{20}) }

V = 1 2800 π = m n π {\bf \implies V = \frac{1}{2800} \pi = \dfrac mn \pi \implies }

m + n = 2801. {\bf m + n = 2801. }

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