Christian's maximum and minimum

Substitute $x = \frac 1 2 \cos t$ and $y = \frac 1 {\sqrt 5} \sin t$ . The expression $2x^2 + 3xy + 2y^2$ then becomes:

$\frac 1 2 \cos^2 t + \frac 2 5 \sin^2 t + \frac 3 {2\sqrt 5} \sin t \cos t$ .

Using the double angle formulae $\cos^2 t = \frac{\cos(2t) + 1} 2$ and $\sin t\cos t = \frac 1 2 \sin 2t$ , the above expression becomes:

$\frac 9 {20} + \frac{\cos 2t} {20} + \frac 3 {4\sqrt 5} \sin 2t$

Now we apply the following trick.

Claim : The maximum and minimum values of $A\cos\theta + B\sin\theta$ are $+\sqrt{A^2+B^2}$ and $-\sqrt{A^2+B^2}$ .

Proof :

By considering the point (A, B) on the circle with radius $R = \sqrt{A^2 + B^2}$ , we can write $A = R\cos\alpha$ and $B = R\sin\alpha$ . Then:

$A \cos\theta + B\sin\theta = R\cos\alpha \cos\theta + R\sin\alpha\sin\theta = R\cos(\theta-\alpha).$

The maximum and minimum values of this expression are clearly +R and -R respectively. (QED)

Conclusion

The maximum and minimum values are $M=\frac 9{20} + \sqrt{\frac {23}{200}}$ and $N=\frac 9{20} - \sqrt{\frac{23}{200}}$ . From this, M+N+MN = 79/80.

I see no obvious inequality method to solve this problem, so we can opt for Lagrange Multipliers . In my honest opinion, this is further motivated by the fact that we want both the extremum . This is what the Lagrange Multipliers say:

tagtext

Denote $g(x,y) = 4x^2 + 5y^2 - 1$ and $f(x,y) = 2x^2 + 3xy + 2y^2$ . So the candidates for the extremal values are the points for which $\nabla g(x,y) = 0$ and the solutions to the system:

• $f_x = \lambda g_x$

• $f_y = \lambda g_y$

• $g(x,y) = 0$

Now firstly, notice that $\nabla g = \langle 8x,10y \rangle$ and $\nabla f = \langle 4x+3y, 3x+4y \rangle$ , we see that the only solution to $\nabla g(x,y) = 0$ is $(x,y) = (0,0)$ which does not satisfy the $g(x,y) = 0$ condition. Whence, we can safely conclude that the critical points are solutions of the system:

• $4x+3y = 8 \lambda x$

• $3x + 4y = 10 \lambda y$

• $4x^2 + 5y^2 = 1$

We isolate variables in the first $2$ equations to get:

$y = ( \frac{8 \lambda}{3}-\frac{4}{3}) x$

$x = ( \frac{10 \lambda}{3} - \frac{4}{3}) y$

Substitute the first equation into the second and we have, $x = ( \frac{10 \lambda}{3} - \frac{4}{3}) ( \frac{8 \lambda}{3}-\frac{4}{3}) x$ so either $x = 0$ or we have to solve the quadratic $( \frac{10 \lambda}{3} - \frac{4}{3}) ( \frac{8 \lambda}{3}-\frac{4}{3}) = 1$ . The former is rejected because of inconsistency within the system.

As for the latter, we solve to get $2$ answers, corresponding to the minimum and maximum respectively: $\lambda = \{ \frac{1}{20} (9+ \sqrt{46}), \frac{1}{20} (9 - \sqrt{46}) \}$ .

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For this final step, we have to notice the important factorisation that $M + MN + N = (M+1)(N+1) - 1$ . Joking, actually, we don't. The more important observation is that we have conjugates in our value of $M ,N$ , so directly adding and multiplying is actually a better idea. So, we have:

$M + MN + N = \frac{1}{20} (9+ \sqrt{46}) + \frac{1}{20} (9+ \sqrt{46})\frac{1}{20} (9 - \sqrt{46}) + \frac{1}{20} (9 - \sqrt{46}) = \frac{18}{20} + \frac{81 - 46}{400} = \frac{79}{80}$ .

In conclusion, the answer that we want is: $a+b = 79+80 = \fbox{159}$ .

Let $P=2x^2+3xy+2y^2$ .

If $y=0 => x^2=\frac{1}{4},P=\frac{1}{2}$

If $y \neq 0$ Let $t=\frac{x}{y}$ and $P=2x^2+3xy+2y^2=\frac{2x^2+3xy+2y^2} {4x^2+5y^2}=\frac{2t^2+3t+2}{4t^2+5}$

Note that $4t^2+5=\frac{1}{y^2}$ So $t \in \mathbb{R}$

Consider the function $f(t)=\frac{2t^2+3t+2}{4t^2+5}; t\in \mathbb{R}$

$f'(x)=\frac{-12t^2+4t+15}{(4t^2+5)^2}$ ; $f'(x)=0 \Leftrightarrow x=\frac{1+\sqrt{46}}{6}=a \vee x=\frac{1-\sqrt{46}}{6}=b$

And so $M=\max P=\max f(x)=f(a)=\frac{9+\sqrt{46}}{20}; N=\min P=\min f(x)= f(b)=\frac{9-\sqrt{46}}{20}$

Then it's easy to calculate the sum $M+N+MN=\frac{79}{80}$

Let Q=2x^2+3xy+2y^2. If y=0=>x2=14,Q=12 If y≠0 Let t=xy and Q=2x^2+3xy+2y^2=(2x^2+3xy+2y^2)/(4x^2+5y^2)=(2t^2+3t+2)/(4t^2+5)

now 4t^2+5=1y^2 So t∈R

Consider the function f(t)=(2t^2+3t+2)/(4t^2+5);

Note that 4t^2+5=1/y^2 t∈R

f′(x)=(−12t^2+4t+15)/(4t^2+5)2;

f′(x)=0⇔x=(1+46)/√6=a∨x=(1−46)/√6=b

And so M=maxQ=maxf(x)=f(a)=(9+46)/√20;N=minQ=minf(x)=f(b)=(9−46)/√20

Then it's easy to calculate the sum M+N+MN=79/80

Let $y=kx$ then $x^2(4+5k^2)=1$ and $A=x^2(2+3k+2k^2)$ Denote $u=\frac{9+sqrt46}{20}$ and $v=\frac{9-sqrt46}{20}$ ( this number comes from solving later equation) Then it is easy to check that $v(4+5k^2)<=2+3k+2k^2<=u(4+5k^2)$ Then the maximum and minimum value of A is u and v. Thus u+v+uv=79/80. the answer is 159.

We will attempt to convert the problem from two variables into one parameter. We can do so using the transformation $(x,y)=(\displaystyle\frac{\cos\theta}{2},\displaystyle\frac{\sin\theta}{\sqrt{5}})$ .

Let us denote the given expression by $P$ .

Using the transformation,

$P=\displaystyle\frac{\cos^2\theta}{2}+\displaystyle\frac{3\sin\theta\cos\theta}{2\sqrt{5}}+\displaystyle\frac{2\sin^2\theta}{5}$

or $P=\displaystyle\frac{3\sin2\theta}{4\sqrt5}+\displaystyle\frac{\cos2\theta}{20}+\displaystyle\frac{9}{20}$ $($ Using double-angle $\sin$ and $\cos$ formulae $)$

Let $A=\sqrt{\big(\displaystyle\frac{3}{4\sqrt5}\big)^2+\big(\displaystyle\frac{1}{20}\big)^2}$ .

Then $M=A+\displaystyle\frac{9}{20},\,N=\,-A+\displaystyle\frac{9}{20}$ .

So $M+N+MN=2\times\displaystyle\frac{9}{20}+(\displaystyle\frac{9}{20})^2-A^2=\displaystyle\frac{79}{80}$ .

Then $a+b=\boxed{159}$ .

Given that $4x^2 + 5y^2 = 1$ $\implies (2x)^2 + (\sqrt 5 y)^2 = 1$ , which is an equation for ellipse. We can substitute $2x = \cos \theta$ and $\sqrt 5 y = \sin \theta$ and then:

$\begin{aligned} 2x^2 + 3xy + 2y^2 & = \frac 12 \cos^2 \theta + \frac 3{2\sqrt 5}\sin \theta \cos \theta + \frac 25 \sin^2 \theta \\ & = \frac 14 (1 + \cos 2 \theta) + \frac 3{4\sqrt 5}\sin 2 \theta + \frac 15 (1-\cos 2 \theta) \\ & = \frac 9{20} + \frac 1{20} \cos 2 \theta + \frac 3{4\sqrt 5}\sin 2 \theta \\ & = \frac 9{20} + \frac {\sqrt{46}}{20} \sin \left(2 \theta + \tan^{-1} \frac {\sqrt 5}{15} \right) \end{aligned}$

Then we have:

$\begin{cases} M = \max \left(\dfrac 9{20} + \dfrac {\sqrt{46}}{20} \sin \left(2 \theta + \tan^{-1} \dfrac {\sqrt 5}{15} \right) \right) = \dfrac {9+\sqrt{46}}{20} & \text{when } \sin \left(2 \theta + \tan^{-1} \dfrac {\sqrt 5}{15} \right) = 1 \\ N = \min \left(\dfrac 9{20} + \dfrac {\sqrt{46}}{20} \sin \left(2 \theta + \tan^{-1} \dfrac {\sqrt 5}{15} \right) \right) = \dfrac {9-\sqrt{46}}{20} & \text{when } \sin \left(2 \theta + \tan^{-1} \dfrac {\sqrt 5}{15} \right) = -1 \end{cases}$

Therefore, $M+N+MN = \dfrac {9+\sqrt{46}}{20} + \dfrac {9-\sqrt{46}}{20} + \dfrac {9+\sqrt{46}}{20} \times \dfrac {9-\sqrt{46}}{20} = \dfrac 9{10} + \dfrac 7{80} = \dfrac {79}{80}$ $\implies a+b = 79+80 = \boxed{159}$ .

Knowing that $4x^2 + 5y^2 = 1$ representes the equation of a ellipsis, is convenient to make the following trigonometric transformations: $$ x = \frac{\cos\theta}{2}, y = \frac{\sin\theta}{\sqrt{5}} $$ Onto the second equation: $$ f(\theta) = 2\cdot\frac{\cos^2\theta}{4} + 3\cdot\frac{\cos\theta}{2}\cdot\frac{\sin\theta}{\sqrt{5}} + 2\cdot\frac{\sin^2\theta}{5} $$ Using double arc transformation and some algebraic manipulation we'll get to: $$ f(\theta) = \frac{1}{20}\cdot\cos(2\theta) + \frac{3}{4\sqrt{5}}\cdot\sin(2\theta) + \frac{9}{20} $$ As we can see, it's always possible to do the following transformation: $$ A\cos x + B\sin x = \sqrt{A^2 + B^2}\left(\frac{A}{\sqrt{A^2 + B^2}}\cos x + \frac{B}{\sqrt{A^2 + B^2}}\sin x\right) $$ It's even possible to see that: $$ A\cos x + B\sin x = \sqrt{A^2 + B^2}\sin(x + \alpha) $$ Where $\alpha$ is obtained contructing a right triangle of sides $A$ and $B$ . Finally, we observe that the maximum and minimum values of $f(\theta)$ occur when $\sin(\alpha + x) = \pm 1$ . Hence: $$ M =\frac{9}{20} + \sqrt{\frac{1}{400} + \frac{9}{80}}, N =\frac{9}{20} - \sqrt{\frac{1}{400} + \frac{9}{80}} $$ Calculating $M + N + MN$ we come to the answer: $$ M + N + MN = \frac{79}{80} \Rightarrow a + b = 159 $$

M = .789116 N = .110884 MN = .087500

M+N+MN = .987500 = 79/80 a+b=79+80 =159

This is very simple ,start off by noticing that 4x^2+5y^2=1 is an equation of an ellipse and the reqn eqn is the equation of a pair of straight line. Now you can use the above to simply solve the prob or u can also use lagrange's theorem . I think ull get M=1/20(9+4root 6 ) and N to be its conjugate

Let $F(x,y)=2x^2+3xy+2y^2$ . We shall consider the following quadratic, where $\lambda$ is a real number: $F(x,y)-\lambda=(2-4\lambda)x^2+3xy+(2-5\lambda)y^2$ We want to find out when we can write this polynomial as $\pm(ax+by)^2$ ; this happens when the discriminant of the quadratic is 0. In other words, $\begin{aligned} 3^2-4(2-4\lambda)(2-5\lambda)&=0\\ 80\lambda^2-72\lambda+7&=0\\ \lambda&=\frac{9\pm\sqrt{46}}{20} \end{aligned}$ When $\lambda=\frac{9+\sqrt{46}}{20}$ , we have $\begin{aligned} F(x,y)-\lambda&=\frac{1-\sqrt{46}}5x^2+3xy+\frac{-1-\sqrt{46}}4\\ &=-\left(\sqrt{\frac{\sqrt{46}-1}5}x+\sqrt{\frac{\sqrt{46}+1}4}y\right)^2\\ &\leq0\end{aligned}$ Hence $F(x,y)\leq\frac{9+\sqrt{46}}{20}$ , and equality holds if and only if $\sqrt{\frac{\sqrt{46}-1}5}x+\sqrt{\frac{\sqrt{46}+1}4}y=0$ and $4x^2+5y^2=1$ . Notice that the first equation describes a line passing through the origin, and the second describes an ellipse centred at the origin. Hence these two simultaneous equations indeed has a solution, and thus there exists $x,y$ such that $F(x,y)=\frac{9+\sqrt{46}}{20}$ . Therefore $\boxed{M=\frac{9+\sqrt{46}}{20}}$ .

Similarly, when $\lambda=\frac{9-\sqrt{46}}{20}$ , we have $\begin{aligned} F(x,y)-\lambda&=\frac{1+\sqrt{46}}5x^2+3xy+\frac{-1+\sqrt{46}}4\\ &=\left(\sqrt{\frac{1+\sqrt{46}}5}x+\sqrt{\frac{-1+\sqrt{46}}4}y\right)^2\\ &\geq0\end{aligned}$ Hence $F(x,y)\geq\frac{9-\sqrt{46}}{20}$ . By a similar argument to the above, equality is possible. Hence $\boxed{N=\frac{9-\sqrt{46}}{20}}$ .

Now $M$ and $N$ are the roots of the quadratic equation $80\lambda^2-72\lambda+7=0$ , so by Vieta's formulas we get $\begin{aligned}M+N&=\frac{72}{80}\\MN&=\frac{7}{80}\\\therefore M+N+MN&=\frac{79}{80},\end{aligned}$ and thus the answer is $79+80=\boxed{159}$ .

M = .789116 N = .110884 MN = .087500

M+N+MN = .987500 = 79/80

Nice...

4x^2+5y^2=1 (1)

A=2x^2 + 3xy + 2y^2 (2)

if y = 0 then x^2= 1/4 => A=1/2

y not equals to 0

(1) <=> 1/y^2= 4 x^2/y^2 + 5

(2) <=> A/y^2= 2 x^2/y^2 + 3x/y +2

<=> A*(4 x^2/y^2 +5)=2 x^2/y^2 + 3x/y +2

<=> (4A-2) x^2/y^2 - 3 x/y +5A -2 =0

9 - 4 (4A-2) (5A-2) >=0

<=> 80A^2 - 72A + 7 <=0 <=> N=.....<= A <= M=....

M+N+MN= 72/80 + 7/80 = 79/80 => a+b = 159 ( N=< A=1/2<=M )

let x and y be described by the parametrization of x=.5cos(t), y=(sin (t))/(sqrt5) This will satisfy the requirement of 4x^{2}+5y^{2}=1. Plugging those into our equation, it now becomes .5(cos(t))^2 + .75/(sqrt5) (sin(2t))+.4(sin(t))^2. Further simplification yields .1(cos (t))^2 + .75/(sqrt5) (sin(2t))+.4. Now, taking the derivative and setting it to zero, we see than tan(2t)=3 sqrt5. That means the maximum and minimum occur when t is in quadrant 1 and quadrant three. In the parametrized equation, the only term that depends on this change is .75/(sqrt5) (sin(2t)). From tan(2t), we can get sin(2t) and cos(2t) from drawing up a triangle, and from there we can get (cos(t))^2 by using the half angle formula. Also, we can rewrite M+N+MN as (M+1)(N+1)-1. Plugging in the values for sin(2t) and (cos(t))^2 will yield the maximum and plugging in -sin(2t) and (cos(t))^2 will yield the minimum. Finally, plugging these values into (M+1)(N+1)-1, we get 79/80. So the answer is 159.

Let Q=2x^2+3xy+2y^2. If y=0=>x2=14,Q=12 If y≠0 Let t=xy and Q=2x^2+3xy+2y^2=(2x^2+3xy+2y^2)/(4x^2+5y^2)=(2t^2+3t+2)/(4t^2+5)

now 4t^2+5=1y^2 So t∈R

Consider the function f(t)=(2t^2+3t+2)/(4t^2+5);

Note that 4t^2+5=1/y^2 t∈R

f′(x)=(−12t^2+4t+15)/(4t^2+5)2;

f′(x)=0⇔x=(1+46)/√6=a∨x=(1−46)/√6=b

And so M=maxQ=maxf(x)=f(a)=(9+46)/√20;N=minQ=minf(x)=f(b)=(9−46)/√20

Then it's easy to calculate the sum M+N+MN=79/80

Let Q=2x^2+3xy+2y^2. If y=0=>x2=14,Q=12 If y≠0 Let t=xy and Q=2x^2+3xy+2y^2=(2x^2+3xy+2y^2)/(4x^2+5y^2)=(2t^2+3t+2)/(4t^2+5)

now 4t^2+5=1y^2 So t∈R

Consider the function f(t)=(2t^2+3t+2)/(4t^2+5);

Note that 4t^2+5=1/y^2 t∈R

f′(x)=(−12t^2+4t+15)/(4t^2+5)2;

f′(x)=0⇔x=(1+46)/√6=a∨x=(1−46)/√6=b

And so M=maxQ=maxf(x)=f(a)=(9+46)/√20;N=minQ=minf(x)=f(b)=(9−46)/√20

Then it's easy to calculate the sum M+N+MN=7980