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Algebra Level 5

$\large x^2(x^2+1)+y^2(y^2+1)$

Positive reals $x$ and $y$ satisfying the system of inequalities below.

$\begin{cases} x\leq y\leq 2 \\ 2x+y\geq 2xy\end{cases}$

Find the maximum value of the expression above.

This problem was taken from HSGS grade 10 selection test

The answer is 22.

2 solutions

P C
Jun 4, 2016

Sorry for posting the wrong answer and solution, here's the correct one. From the condition, we have $\frac{1}{x}+\frac{2}{y} \ge 2$ Set $a=\frac{1}{x}, b=\frac{1}{y}$ , the expression becomes $\frac{1}{a^4}+\frac{1}{a^2}+\frac{1}{b^4}+\frac{1}{b^2}$ and the conditions are $\begin{cases} a+2b\geq 2 \\ a\geq b\geq\frac{1}{2}\end{cases}$ By C-S Inequality we get $\begin{cases}a^2+(2b)^2\geq 2\\ a^4+(2b)^4\geq 2\end{cases}$ . Now we see that $\bigg(1-\frac{b^2}{a^2}\bigg)\bigg(\frac{1}{b^2}-4\bigg)\leq 0\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}\leq\frac{2-4b^2}{a^2}+4\leq 5$ $\bigg(1-\frac{b^4}{a^4}\bigg)\bigg(\frac{1}{a^4}-16\bigg)\leq 0\Rightarrow \frac{1}{a^4}+\frac{1}{b^4}\leq 17$ Therefore the maximum value is 22. The equality holds when $(x,y)=(1;2)$

But if we take point (1; 2) which also satisfies the conditions we get a larger value (=22) for the given expression!

Andreas Wendler - 5 years ago

I've fixed the solution, you can check it back

P C - 5 years ago

ah, yes, tks for pointing it out, please report it so the staff can edit the answer

P C - 5 years ago

Can you explain the first line? Note that you divided by $2 (y-1)$ which could potentially be negative.

Calvin Lin Staff - 5 years ago

I'm trying to come up with a new solution because this one is really loose

P C - 5 years ago

So here's one possible way of fixing it: We have $f(1,2) = 22$ . Suppose we want to find a value larger than that. Then, since $x \leq y \leq 2$ if $x < 1$ we will have $f(x,y) < f(1,2)$ . Hence, we may include the constraint that $x \geq 1$ . This gives us $1 \leq x \leq y \leq 2$ , and so $y-1$ is non-negative. We can thus divide by that value, and proceed.

Having said that, note that line 2 is not true. If we take $y\rightarrow 1^+$ then clearly it goes towards infinity. Hence, there is no upper bound that we can set on the expression.

Calvin Lin Staff - 5 years ago

@Calvin Lin – I've posted another solution, this's the solution of Vo Quoc Ba Can, not mine

P C - 5 years ago

@P C – Can you check your signs? You show that one term is less than 5 and the other is greater than 17, and conclude that the sum is less than 22. Also, it's missing an inequality sign in the last statement.

Calvin Lin Staff - 5 years ago

I get 22 , Only use my logic

Novril Razenda - 4 years, 11 months ago

ho did you get (2x-1)(1-y)>-1

abhishek alva - 5 years ago

A little problem... If y=1, we have 2x + 1 ≥ 2x. So if x=-∞, then x²(x²+1) + y²(y²+1) = +∞. So we have a problem with conditions...

Abysse Synus - 5 years ago

Actually, $x$ and $y$ are positive reals.

Manuel Kahayon - 5 years ago

According to ineq.(I) max. Value of x = y Put the value in ineq.(II) 2x+x>= 2x.x X<= 3/2; max value of x = 3/2 = y who satisfied both ineq. 9/4(9/4+1)+9/4(9/4+1) =117/8; [117/8] = 14

Vivek Kumar Gupta - 5 years ago

I've try that approach and got the wrong answer

P C - 5 years ago

Mayank Jha
Jun 13, 2016

The best way to quickly get to the solution is to use inequality 2 to get maximum values of x and y subject to inequality 1 that x = 1,y=2.(2x=y=2xy/2).Then put the values of x and y in the expression given above to get the maximum value of the expression to be 22.

Are you smarter than a grade 9 student?

This problem was taken from HSGS grade 10 selection test

The answer is 22.

2 solutions

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