For positive reals x and y satisfying 3 x + 4 y = 5 . If the maximum value of x 2 y 3 can be expressed as B A for coprime positive integers A and B , find the value of A + B .
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Nice solution...
I also applied same method. ;)
Did the exact same
Used the same approach. :D
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 . 1 8 7 5 . Here's the way:-
5 2 x + x + 2 y + y + y ≥ 5 ( 2 x 2 ) ( 2 y 3 ) 4 1 ≥ y 3 x 2
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I think that that method is wrong because the maxima will then occur at 2x=x=2y=y.
Your method is incorrect . You have to break in equal parts.
Let x = a / 3 and y = b / 4 , then we have b = 5 − a and we maximize f ( a ) = 3 2 ⋅ 4 3 a 2 ( 5 − a ) 3 .
The maximum corresponds to a derivative of zero: f ′ ( a ) = 3 2 ⋅ 4 3 1 ⋅ ( 2 a ( 5 − a ) 3 − 3 a 2 ( 5 − a ) 2 ) = 3 2 ⋅ 4 3 1 ⋅ ( 2 ( 5 − a ) − 3 a ) ⋅ a ( 5 − a ) 2 = 0 . The only non-trivial solution (which corresponds to a maximum) is 2 ( 5 − a ) − 3 a = 0 or a = 2 .
Finishing up: b = 5 − a = 3 , and f ( a ) = 3 2 ⋅ 4 3 2 2 ⋅ 3 3 = 4 2 3 , so that the solution is 3 + 1 6 = 1 9 .
This approach is much more easy and convinient.
Oh! I got it. Thank you so much . 😊
Derivative also yields same result
Could you please explain it using derivative
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I posted a solution with derivatives.
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I am unable to find it
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@Manvendra Singh – Sorry, I forgot to hit "Publish". I should be up there now.
Please mention that x and y are positive real numbers
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We note that: 3 x + 4 y = 2 3 x + 2 3 x + 3 4 y + 3 4 y + 3 4 y and that:
2 3 x + 2 3 x + 3 4 y + 3 4 y + 3 4 y ⇒ 5 1 ⇒ x 2 y 3 ≥ 5 5 ( 2 3 x ) 2 ( 3 4 y ) 3 ≥ 5 5 3 1 6 x 2 y 3 ≥ 3 1 6 x 2 y 3 ≤ 1 6 3 = 0 . 1 8 7 5
Note that the same result is obtained using Lagrange multiplier method, which does not need x and y to be positive.
Using Lagrange Multiplier – as requested by Shivam Mishra
Let F ( x , y , λ ) = x 2 y 3 − λ ( 3 x + 4 y − 5 ) , where λ is the Lagrange multiplier and 3 x + 4 y − 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 ≤ ( 3 2 ) 2 ( 4 3 ) 3 = 1 6 3 ⇒ A + B = 3 + 1 6 = 1 9