Olympiad Corner 2

Algebra Level 4

For positive reals x x and y y satisfying 3 x + 4 y = 5 3x + 4y = 5 . If the maximum value of x 2 y 3 x^{2}y^{3} can be expressed as A B \dfrac AB for coprime positive integers A A and B B , find the value of A + B A+B .


The answer is 19.

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4 solutions

Chew-Seong Cheong
Nov 26, 2015

We note that: 3 x + 4 y = 3 2 x + 3 2 x + 4 3 y + 4 3 y + 4 3 y 3x + 4y = \dfrac{3}{2}x + \dfrac{3}{2}x + \dfrac{4}{3}y + \dfrac{4}{3}y + \dfrac{4}{3}y and that:

3 2 x + 3 2 x + 4 3 y + 4 3 y + 4 3 y 5 ( 3 2 x ) 2 ( 4 3 y ) 3 5 5 5 16 x 2 y 3 3 5 1 16 x 2 y 3 3 x 2 y 3 3 16 = 0.1875 \begin{aligned} \frac{3}{2}x + \frac{3}{2}x + \frac{4}{3}y + \frac{4}{3}y + \frac{4}{3}y & \ge 5 \sqrt[5]{\left(\frac{3}{2}x\right)^2 \left(\frac{4}{3}y\right)^3} \\ \Rightarrow 5 & \ge 5 \sqrt[5]{\frac{16x^2y^3}{3}} \\ 1 & \ge \frac{16x^2y^3}{3} \\ \Rightarrow x^2y^3 & \le \frac{3}{16} = \boxed{0.1875} \end{aligned}

Note that the same result is obtained using Lagrange multiplier method, which does not need x x and y y to be positive.

Using Lagrange Multiplier – as requested by \color{#3D99F6}{\text{Using Lagrange Multiplier -- as requested by }} Shivam Mishra

Let F ( x , y , λ ) = x 2 y 3 λ ( 3 x + 4 y 5 ) F(x,y,\lambda) = x^2y^3 - \lambda(3x+4y - 5) , where λ \lambda is the Lagrange multiplier and 3 x + 4 y 5 3x+4y-5 is the constraint. Then, we have:

\(\begin{array} {} \dfrac{\partial F}{\partial x} = 0 & \Rightarrow 2xy^3 = 3 \lambda & ... (1) \\ \dfrac{\partial F}{\partial y} = 0 & \Rightarrow 3x^2y^2 = 4 \lambda & ... (2) \\ \dfrac{\partial F}{\partial \lambda} = 0 & \Rightarrow 3x+4y=5 & ... (3) \end{array} \)

\(\begin{array} {} \dfrac{(1)}{(2)}: & \dfrac{2y}{3x} = \dfrac{3}{4} & \Rightarrow y = \dfrac{9}{8}x \\ (3): & 3x + 4 \times \dfrac{9}{8}x = 5 & \Rightarrow x = \frac{2}{3} & \Rightarrow y = \frac{3}{4} \end{array} \)

x 2 y 3 ( 2 3 ) 2 ( 3 4 ) 3 = 3 16 A + B = 3 + 16 = 19 \Rightarrow x^2y^3 \le \left(\frac{2}{3}\right)^2 \left(\frac{3}{4}\right)^3 = \frac{3}{16} \\ \Rightarrow A + B = 3+16 = \boxed{19}

Nice solution...

Dev Sharma - 5 years, 6 months ago

I also applied same method. ;)

Priyanshu Mishra - 5 years, 6 months ago

Did the exact same

Aditya Kumar - 5 years ago

Used the same approach. :D

A Former Brilliant Member - 5 years, 6 months ago

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ye me too same approach

Kaustubh Miglani - 5 years, 6 months ago

awesome way I did the same but I believe there is something that is messing up, I also got a maximum which was greater than 0.1875 0.1875 . Here's the way:-

2 x + x + 2 y + y + y 5 ( 2 x 2 ) ( 2 y 3 ) 5 1 4 y 3 x 2 \Large{\frac { 2x+x+2y+y+y }{ 5 } \ge \sqrt [ 5 ]{ \left( 2{ x }^{ 2 } \right) \left( 2{ y }^{ 3 } \right) } \\ \frac { 1 }{ 4 } \ge y^{ 3 }{ x }^{ 2 }}

Department 8 - 5 years, 6 months ago

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I think that that method is wrong because the maxima will then occur at 2x=x=2y=y.

Harshit Singhania - 5 years, 6 months ago

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Oh thanks did not saw that

Department 8 - 5 years, 6 months ago

Your method is incorrect . You have to break in equal parts.

Rahul Kamble - 5 years, 6 months ago
Arjen Vreugdenhil
Nov 27, 2015

Let x = a / 3 x = a/3 and y = b / 4 y = b/4 , then we have b = 5 a b = 5 - a and we maximize f ( a ) = a 2 ( 5 a ) 3 3 2 4 3 . f(a) = \frac {a^2\ (5-a)^3}{3^2\cdot 4^3}.

The maximum corresponds to a derivative of zero: f ( a ) = 1 3 2 4 3 ( 2 a ( 5 a ) 3 3 a 2 ( 5 a ) 2 ) = 1 3 2 4 3 ( 2 ( 5 a ) 3 a ) a ( 5 a ) 2 = 0. f'(a) = \frac1{3^2\cdot 4^3}\cdot \left(2a(5-a)^3-3a^2(5-a)^2\right) = \frac 1{3^2\cdot 4^3}\cdot \left(2(5-a)-3a\right)\cdot a(5-a)^2 = 0. The only non-trivial solution (which corresponds to a maximum) is 2 ( 5 a ) 3 a = 0 2(5-a)-3a=0 or a = 2 a = 2 .

Finishing up: b = 5 a = 3 b = 5 - a = 3 , and f ( a ) = 2 2 3 3 3 2 4 3 = 3 4 2 , f(a) = \frac{2^2\cdot 3^3}{3^2 \cdot 4^3} = \frac{3}{4^2}, so that the solution is 3 + 16 = 19 3 + 16 = \boxed{19} .

This approach is much more easy and convinient.

Manvendra Singh - 5 years, 6 months ago

Oh! I got it. Thank you so much . 😊

Manvendra Singh - 5 years, 6 months ago
Gopal Narayanan
Nov 26, 2015

Derivative also yields same result

Could you please explain it using derivative

Manvendra Singh - 5 years, 6 months ago

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I posted a solution with derivatives.

Arjen Vreugdenhil - 5 years, 6 months ago

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I am unable to find it

Manvendra Singh - 5 years, 6 months ago

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@Manvendra Singh Sorry, I forgot to hit "Publish". I should be up there now.

Arjen Vreugdenhil - 5 years, 6 months ago
Aakash Khandelwal
Nov 25, 2015

Please mention that x and y are positive real numbers

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