Tricky Maxima..!!!

Algebra Level 4

Over all sets of real numbers ( x , y ) (x,y) such that x 2 + y 2 = 1 { x }^{ 2 }+{ y }^{ 2 }=1 , find the maximum value of F ( x , y ) = ( 3 x + 4 y ) 2 + ( 8 x + 6 y ) 2 F(x,y)= ({ 3x+4y })^{ 2 } + { (8x+6y) }^{ 2 }

to 2 decimal places

This is part of this set


The answer is 123.41.

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Deepanshu Gupta by Deepanshu Gupta , 25, India

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5 solutions

Since x 2 + y 2 = 1 x^2+y^2 = 1 , we can let x = cos ⁡ θ x = \cos \theta and y = sin ⁡ θ y = \sin \theta . Therefore, we have:

F ( x , y ) = ( 3 x + 4 y ) 2 + ( 8 x + 6 y ) 2 = 73 x 2 + 120 x y + 52 y 2 As x 2 + y 2 = 1 = 21 x 2 + 120 x y + 52 = 21 cos ⁡ 2 θ + 120 cos ⁡ θ sin ⁡ θ + 52 = 21 2 ( cos ⁡ ( 2 θ ) + 1 ) + 60 sin ⁡ ( 2 θ ) + 52 = 10.5 cos ⁡ ( 2 θ ) + 60 sin ⁡ ( 2 θ ) + 62.5 = 10. 5 2 + 6 0 2 sin ⁡ ( 2 θ + ϕ ) + 62.5 where ϕ = tan ⁡ − 1 10.5 60 \begin{aligned} F(x,y) & = (3x+4y)^2 + (8x + 6y)^2 \\ & = 73x^2 +120xy +52y^2 \quad \quad \small \color{#3D99F6}{\text{As } x^2+y^2 = 1} \\ & = 21x^2 +120xy +52 \\ & = 21\cos^2 \theta +120\cos \theta \sin \theta +52 \\ & = \frac{21}{2}(\cos (2 \theta) +1 ) +60 \sin (2 \theta) +52 \\ & = 10.5 \cos (2 \theta) + 60 \sin (2 \theta)+62.5 \\ & = \sqrt{10.5^2 +60^2} \sin (2 \theta + \color{#3D99F6}{\phi})+62.5 \quad \quad \small \color{#3D99F6}{\text{where } \phi = \tan^{-1} \frac{10.5}{60}} \end{aligned}

Since maximum of sin ⁡ ( 2 θ + ϕ ) = 1 \sin (2 \theta +\phi) = 1 , then we have:

F ( x , y ) m a x = 10. 5 2 + 6 0 2 + 62.5 = 60.91182151 + 62.5 ≈ 123.41 \begin{aligned} F(x,y)_{max} & = \sqrt{10.5^2 +60^2}+62.5 \\ & = 60.91182151 + 62.5 \\ & \approx \boxed{123.41} \end{aligned}

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Nice approach.(+1.

Niranjan Khanderia - 2 years, 11 months ago
Abhishek Sinha
Aug 19, 2014

Consider the linear transformation A = ( 3 4 8 6 ) A=\begin{pmatrix} 3 & 4\\ 8& 6\end{pmatrix} . By performing Singular Value Decomposition of this matrix, we find that the maximum singular value of the matrix A A is 11.1091 11.1091 . Thus the maximum value of F ( x , y ) F(x,y) is simply the Rayleigh quotient of A T A A^{T}A , which is given by the square of the maximum singular value of A A , i.e., 11.109 1 2 ≈ 123.41 11.1091^2\approx 123.41 .

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Ankit Akash
Sep 5, 2014

We substitute x=sinß to get y=cosß from 1st relation. Now we write see second equation in form of sin2ß and cos 2ß and apply Cauchy Schwartz to get the answer

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did the same way!

Mvs Saketh - 6 years, 3 months ago

exactly what i did

Mayank Singh - 6 years, 2 months ago
Hansraj Sharma
Aug 18, 2014

here is a tricky method first we take x=0,y=1 then F(x,y)=16+36=52 second we take x=1,y=0 then F(x,y)=9+64=73 after that 52+73=125........got it

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Moderator note:

This solution is incorrect. There is no justification, nor is it correct, that max ⁡ F ( x , y ) = F ( 0 , 1 ) + F ( 1 , 0 ) \max F(x,y) = F(0,1) + F(1,0) .

Why did u added the two values ???? That will give value of 2f(x,y)...

Sandeep Bhardwaj - 6 years, 9 months ago

The elegant solution is :: as give x^{2} +y^{2} = 1 ..we can assume x=sin(a) and y=cos(a)..so max value of 3sin(a) + 4cos(a) = 5. and max value of 8sin(a) + 6cos(a) =10. Hence max value of f(x,y)=5^2 + 10^2 = 125.

Sandeep Bhardwaj - 6 years, 9 months ago

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This is not justified as the two expressions attain maximum values at different values of a a , namely a = 2 tan ⁡ − 1 1 3 a=2\tan^{-1} {\dfrac{1}{3}} and a = 2 tan ⁡ − 1 1 2 a=2\tan^{-1} {\dfrac{1}{2}} .

The correct approach would be to maximize 21 2 ( cos ⁡ 2 a − sin ⁡ 2 a ) + 120 sin ⁡ a cos ⁡ a + 125 2 = 1 2 ( 120 sin ⁡ 2 a + 21 cos ⁡ 2 a + 125 ) \dfrac{21}{2} (\cos^2 a-\sin^2 a) +120\sin a \cos a+\dfrac{125}{2}\\ =\dfrac{1}{2} (120\sin 2a+21\cos 2a+125)

Hence, the maximum value of the expression would be max ⁡ { F ( x , y ) } = 12 0 2 + 2 1 2 + 125 2 = 3 1649 + 125 2 ≈ 123.4118 \begin{aligned} \max\{F(x,y)\} &=\dfrac{\sqrt{120^2+21^2}+125}{2}\\ &=\dfrac{3\sqrt {1649}+125}{2}\\ &\approx \boxed{123.4118} \end{aligned}

So the answer provided is wrong.

Pratik Shastri - 6 years, 9 months ago

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I too have done the same thing and got the same answer

Ronak Agarwal - 6 years, 9 months ago

first i find 123.412 but that was wrong so i take a tricky formula. it is just tricky not real.

hansraj sharma - 6 years, 9 months ago

L e t F ( x , y ) = ( 3 x + 4 y ) 2 + 4 ∗ ( 4 x + 3 y ) 2 L e t F ( m , n ) b e t h e p o i n t w h e r e F i s m a x i m u m , w i t h c o n d i t i o n t h a t i t i s a l s o o n x 2 + y 2 = 1. ∴ t h e y h a v e c o m m o n t a n g e n t a t ( m , n ) . ⟹ d y d x m , n f r o m F ( x , y ) = 0 a n d x 2 + y 2 = 1 , a r e e q u a l . 2 x + 2 y d y d x = 0. ⟹ d y d x = − x y . A n d d F d x = 2 ( 3 x + 4 y ) ∗ ( 3 + 4 d y d x ) + 8 ∗ ( 4 x + 3 y ) ∗ ( 4 + 3 d y d x ) = 0. B u t d y d x = − x y , s u b s t i t u t i n g d y d x b y − x y , a n d m u l t i p l y i n g b y y 2 , d F d x = ( 3 x + 4 y ) ∗ ( 3 y − 4 x ) + 4 ∗ ( 4 x + 3 y ) ∗ ( 4 y − 3 x ) = 0. ∴ − 12 x 2 − 7 x y + 12 y 2 − 48 x 2 + 28 x y + 48 y 2 = − 60 x 2 + 21 x y + 60 y 2 = 0. ⟹ 20 ( y x ) 2 + 7 ( y x ) − 20 = 0 S o l v i n g q u a d r a t i c i n y x , y x = 0.840197............. ( A ) ∴ y 2 x 2 = 0.705931 1 . ⟹ y 2 x 2 + y 2 = y 2 1 = 0.705931 1.705931 . ∴ y = 0.705931 1.705931 = 0.64328. ∴ b y ( A ) x = 0.64328 0.840197 = 0.76563. S o ( m , n ) = ( 0.76563 , 0.64328 ) . ∴ F ( m , n ) = 123.437. Let~~F(x,y)=(3x+4y)^2+4*(4x+3y)^2\\ ~~~~\\ Let~F(m,n)~be~the~~point~where~F~is~maximum,~with ~condition~that~ it~ is~also~on~x^2+y^2=1.\\ \therefore~~ they~ have~ common~ tangent ~at~(m,n).~~~\implies~~\dfrac{dy}{dx}_{m,n}~from~F(x,y)=0~~and~~x^2+y^2=1, ~are~equal.\\ ~~~\\ 2x+2y\dfrac{dy}{dx}=0.~~\implies~\dfrac{dy}{dx}=-\dfrac x y.\\ And~~\dfrac{dF}{dx}=2(3x+4y)*(3+4\dfrac{dy}{dx})+ 8*(4x+3y)*(4+3\dfrac{dy}{dx})=0.\\ ~~~~\\ ~~~\\ But~\dfrac{dy}{dx}=-\dfrac x y,~~substituting~\dfrac{dy}{dx}~~by~~-\dfrac x y,~~and ~multiplying~by~\dfrac y 2,\\ \dfrac{dF}{dx}= (3x+4y)*(3y-4x)+ 4*(4x+3y)*(4y-3x)=0.\\ \therefore~-12x^2-7xy+12y^2-48x^2+28xy+48y^2= -60x^2+21xy+60y^2=0.\\ \implies~~20(\dfrac y x)^2+7(\dfrac y x)-20=0\\ Solving~quadratic~in~\dfrac y x,~~~~~\dfrac y x =0.840197.............(A)\\ ~~~~~~\\ ~~~\\ \therefore~~~~\dfrac {y^2}{ x^2}=\dfrac{0.705931} 1.\\ \implies~ \dfrac {y^2}{ x^2+y^2}=\dfrac {y^2} 1 = \dfrac {0.705931}{ 1.705931}.\\ \therefore~~~y=\sqrt{\dfrac {0.705931}{ 1.705931} }=0.64328.\\ \therefore~by~(A)~~x=\dfrac{0.64328}{0.840197}=0.76563.\\ ~~~\\ ~~~~\\ So~~(m,n)=(0.76563,0.64328).\\ \therefore~~F(m,n)=\Huge~~\color{#D61F06}{123.437}.

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L e t F ( x , y ) = ( 3 x + 4 y ) 2 + 4 ∗ ( 4 x + 3 y ) 2 L e t F ( m , n ) b e t h e p o i n t w h e r e F i s m a x i m u m , w i t h c o n d i t i o n t h a t i t i s a l s o o n x 2 + y 2 = 1. ∴ t h e y h a v e c o m m o n t a n g e n t a t ( m , n ) . ⟹ d y d x m , n f r o m F ( x , y ) = 0 a n d x 2 + y 2 = 1 , a r e e q u a l . 2 x + 2 y d y d x = 0. ⟹ d y d x = − x y . A n d d F d x = 2 ( 3 x + 4 y ) ∗ ( 3 + 4 d y d x ) + 8 ∗ ( 4 x + 3 y ) ∗ ( 4 + 3 d y d x ) = 0. B u t d y d x = − x y , s u b s t i t u t i n g d y d x b y − x y , a n d m u l t i p l y i n g b y y 2 , d F d x = ( 3 x + 4 y ) ∗ ( 3 y − 4 x ) + 4 ∗ ( 4 x + 3 y ) ∗ ( 4 y − 3 x ) = 0. ∴ − 12 x 2 − 7 x y + 12 y 2 − 48 x 2 + 28 x y + 48 y 2 = − 60 x 2 + 21 x y + 60 y 2 = 0. ⟹ 20 ( y x ) 2 + 7 ( y x ) − 20 = 0 S o l v i n g q u a d r a t i c i n y x , y x = 0.840197............. ( A ) ∴ y 2 x 2 = 0.705931 1 . ⟹ y 2 x 2 + y 2 = y 2 1 = 0.705931 1.705931 . ∴ y = 0.705931 1.705931 = 0.64328. ∴ b y ( A ) x = 0.64328 0.840197 = 0.76563. S o ( m , n ) = ( 0.76563 , 0.64328 ) . ∴ F ( m , n ) = 123.437. Let~~F(x,y)=(3x+4y)^2+4*(4x+3y)^2\\ ~~~~\\ Let~F(m,n)~be~the~~point~where~F~is~maximum,~with ~condition~that~ it~ is~also~on~x^2+y^2=1.\\ \therefore~~ they~ have~ common~ tangent ~at~(m,n).~~~\implies~~\dfrac{dy}{dx}_{m,n}~from~F(x,y)=0~~and~~x^2+y^2=1, ~are~equal.\\ ~~~~\\ ~~~\\ 2x+2y\dfrac{dy}{dx}=0.~~\implies~\dfrac{dy}{dx}=-\dfrac x y.\\ And~~\dfrac{dF}{dx}=2(3x+4y)*(3+4\dfrac{dy}{dx})+ 8*(4x+3y)*(4+3\dfrac{dy}{dx})=0.\\ ~~~~\\ ~~~\\ But~\dfrac{dy}{dx}=-\dfrac x y,~~substituting~\dfrac{dy}{dx}~~by~~-\dfrac x y,~~and ~multiplying~by~\dfrac y 2,\\ \dfrac{dF}{dx}= (3x+4y)*(3y-4x)+ 4*(4x+3y)*(4y-3x)=0.\\ \therefore~-12x^2-7xy+12y^2-48x^2+28xy+48y^2= -60x^2+21xy+60y^2=0.\\ \implies~~20(\dfrac y x)^2+7(\dfrac y x)-20=0\\ Solving~quadratic~in~\dfrac y x,~~~~~\dfrac y x =0.840197.............(A)\\ ~~~~~~\\ ~~~\\ \therefore~~~~\dfrac {y^2}{ x^2}=\dfrac{0.705931} 1.\\ \implies~ \dfrac {y^2}{ x^2+y^2}=\dfrac {y^2} 1 = \dfrac {0.705931}{ 1.705931}.\\ \therefore~~~y=\sqrt{\dfrac {0.705931}{ 1.705931} }=0.64328.\\ \therefore~by~(A)~~x=\dfrac{0.64328}{0.840197}=0.76563.\\ ~~~\\ ~~~~\\ So~~(m,n)=(0.76563,0.64328).\\ \therefore~~F(m,n)=\Huge~~\color{#D61F06}{123.437}.

Niranjan Khanderia - 2 years ago

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