Nice use of AM GM

Calculus Level 5

We are given 4 lines with equations:

  • y = ( 2 + 3 a 2 + 3 b 2 ( 1 + a 2 ) 2 + b ( 6 a + 6 a 3 ) a + b ( 1 + a 2 ) ) min ⁡ 2 y = \Bigg( \dfrac{2 + 3a^2 + 3b^2(1 + a^2)^2 + b(6a + 6a^3)}{a + b(1 + a^2)} \Bigg)^2_{\min}
  • y = ( c 2 d 2 − 2 c 2 d + 2 c 2 + 2 c d − 2 c + 1 c 2 d + c ) min ⁡ + 3 − 8 y = \Bigg(\dfrac{c^2d^2 - 2c^2d + 2c^2 + 2cd - 2c + 1}{c^2d + c} \Bigg)_{\min} + 3 - \sqrt{8}
  • y = [ e ] x 2 y = [e]x^2
  • y = 1 2 [ e ] x 2 y =\dfrac{1}{2} [e]x^2

With a , b , c , d > 0 , e ≥ 1 a,b,c,d > 0, e\geq 1 .

The area bounded by the given lines and the curves when obtain a maximum value of M M , with M = a 1 a 2 ( a 3 − 1 ) ( a 4 a 5 − 1 ) M = \dfrac{a_{1}}{a_{2}}(\sqrt{a_{3}} - 1)(a_{4}\sqrt{a_{5}} - 1) .

Find a 1 + a 2 + a 3 + a 4 + a 5 a_{1} + a_{2} + a_{3} + a_{4} + a_{5} .

Details and Assumptions :

  • [ . ] [.] denote the greatest integer function.
  • ( f ( x ) ) min (f(x))_{\text{min}} denote the minimum value of f ( x ) f(x) .
Dedicated to Aditya Raut and Guilherme Dela Corte .


The answer is 63.

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U Z by U Z , 24, Guyana

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1 solution

U Z
Dec 31, 2014

Line 1

2 + 3 y 2 + 3 x 2 y 2 ( 1 y + y ) 2 + 6 x y 2 ( y + 1 y ) y [ 1 + x ( 1 y + y ) ] \dfrac{2 + 3y^{2} + 3x^{2}y^{2}(\frac{1}{y} + y)^{2} + 6xy^{2}(y + \frac{1}{y})}{y[1 + x(\frac{1}{y} + y)]}

= 3 y 2 ( x 2 ( y + 1 y ) 2 + 2 x ( 1 y + y ) + 1 ) y [ 1 + x ( 1 y + y ) ] + 2 y [ 1 + x ( 1 y + y ) ] = \dfrac{3y^{2}( x^{2}(y + \frac{1}{y})^{2} + 2x(\dfrac{1}{y} + y) + 1)}{y[1 + x(\dfrac{1}{y} + y)]} + \dfrac{2}{y[1 + x(\dfrac{1}{y} + y)]}

= 3 y ( x ( 1 y + y ) + 1 ) + 2 y [ 1 + x ( 1 y + y ) ] = 3y( x(\dfrac{1}{y} + y) + 1) + \dfrac{2}{y[1 + x(\dfrac{1}{y} + y)]}

A . M ≥ G . M A.M \geq G.M

= ( 2 × 6 ) 2 = 24 = (2 \times \sqrt{6})^2 = 24 , y = 24 y = 24

Line 2 x 2 y 2 + 2 x y + 1 − 2 x 2 y + 2 x 2 − 2 x x ( x y + 1 ) \dfrac{ x^{2}y^{2} + 2xy + 1 - 2x^{2}y + 2x^{2} - 2x}{x(xy + 1)}

= ( x y + 1 ) 2 − 2 x ( x y + 1 − x ) x ( x y + 1 ) = \dfrac{ (xy + 1)^{2} - 2x(xy + 1 - x)}{x(xy + 1)}

= x y + 1 x + 2 x x y + 1 − 2 = \dfrac{xy + 1}{x} + \dfrac{2x}{xy + 1} - 2

A . M ≥ G . M A.M \geq G.M

x y + 1 x + 2 x x y + 1 2 ≥ 2 \dfrac{\dfrac{xy + 1}{x} + \dfrac{2x}{xy + 1}}{2} \geq \sqrt{2}

m i n i m u m v a l u e = 2 2 − 2 + 3 − 2 2 minimum ~value = 2\sqrt{2} - 2 + 3 - 2\sqrt{2} , y = 1 y = 1

A r e a = 2 ∫ y = 1 y = 24 ( 2 y [ a ] − y [ a ] ) d y Area~ = \displaystyle 2\int_{y=1}^{y=24} (\sqrt{\dfrac{2y}{[a]}} - \sqrt{\dfrac{y}{[a]}})dy

Area will be maximum when [ e ] = 1 [e] = 1

A r e a = 2 ( 2 − 1 ) ∫ 1 24 y d y Area~ = \displaystyle 2(\sqrt{2} - 1) \int_{1}^{24} \sqrt{y}dy

= 4 ( 2 − 1 ) 3 . ( 2 4 3 / 2 − 1 ) = \dfrac{4(\sqrt{2} - 1)}{3}.(24^{3/2} - 1)

A r e a = 4 ( 2 − 1 ) 3 ( 48 6 − 1 ) \boxed{Area ~ = \dfrac{4(\sqrt{2} - 1)}{3}(48\sqrt{6} - 1)}

14 Helpful 0 Interesting 0 Brilliant 0 Confused

The question was Awesome! (Thumbs up)

Aman Gautam - 6 years, 5 months ago

Line 2 - This problem was earlier posted by @Guilhereme Dela Corte

Line 1 - This question appeared in JOMO 10 created by @Aditya Raut

U Z - 6 years, 5 months ago

Loved this problem !! : ) :)

Keshav Tiwari - 5 years, 8 months ago

Ok, manca qualche precisazione su a1 a2 etc

Claudia Manotti - 4 years, 9 months ago

Excellent question. However, the solution would have been clearer had you used a , b , c , d a,b,c,d to determine Lines 1 and 2, in stead of using x , y x,y and causing confusion.

Janardhanan Sivaramakrishnan - 4 years, 8 months ago

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