2 Equations, 5 Variables

Algebra Level 4

x 1 , x 2 , x 3 , x 4 x_1, x_2, x_3, x_4 and x 5 x_5 are reals such that x 1 + x 2 + x 3 + x 4 + x 5 = 8 x_1+x_2+x_3+x_4+x_5=8 and x 1 2 + x 2 2 + x 3 2 + x 4 2 + x 5 2 = 16. x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=16. What is the largest possible value of x 5 ? x_5?


The answer is 3.2.

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8 solutions

Satvik Golechha
Sep 10, 2014

From the given equations, x 1 + x 2 + x 3 + x 4 = 8 x 5 x_1+x_2+x_3+x_4=8-x_5 and x 1 2 + x 2 2 + x 3 2 + x 4 2 = 16 x 5 2 x_1^2 + x_2^2 + x_3^2 + x_4^2=16-x_5^2 .

Apply Cauchy-Schwarz to get:

( x 1 2 + x 2 2 + x 3 2 + x 4 2 ) ( 1 + 1 + 1 + 1 ) ( x 1 + x 2 + x 3 + x 4 ) 2 (x_1^2+x_2^2+x_3^2+x_4^2)(1+1+1+1) \geq (x_1+x_2+x_3+x_4)^2

So, 4 ( 16 x 5 2 ) ( 8 x 5 ) 2 4(16-x_5^2) \geq (8-x_5)^2 , which reduces to 16 5 x 5 0 \frac{16}{5} \geq x_5 \geq 0 .

The equality case occurs where x 1 = x 2 = x 3 = x 4 = 1.2 x_1=x_2=x_3=x_4=1.2 and x 5 = 3.2 . x_5=\boxed{3.2}.

This is a familiar AIME , RMO, and Brilliant problem.

In the question it is given real numbers not positive real.....then how can you apply Cauchy Schwartz.....

Ankur Verma - 3 years ago

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CS is true for all reals.

Satvik Golechha - 3 years ago

can u do it with TITU lemma

Anshaj Shukla - 2 years, 10 months ago

This answer is really awesome ..nice

Magnas Bera - 1 year, 11 months ago

I solved it is a completely different way, but I made a mistake in counting and didn't answer correctly. I just thought to post my solution anyway in a comment. The locus x 1 2 + x 2 2 + x 3 2 + x 5 2 = 16 x_1^2 + x_2^2 + x_3^2 + x_5^2 = 16 is a 4th-dimensional sphere in R 5 \mathbb R^5 . The locus x 1 + x 2 + x 3 + x 4 + x 5 = 8 x_1 + x_2 + x_3 + x_4 + x_5 = 8 is an iper-plane and a very special one in fact each unit vector of the standard base (1,0,0,0,0), (0,1,0,0,0),(0,0,1,0,0) (0,0,0,1,0) and (0,0,0,0,1) forms with the normal vector of the plane (that is (1,1,1,1,1) btw) an angle of 45 degrees. The intersection of the plane and the sphere is a 3rd-dimensional sphere centered in C = ( 8 / 5 , 8 / 5 , 8 / 5 , 8 / 5 , 8 / 5 ) C = (8/5, 8/5, 8/5, 8/5, 8/5) and with radius 8 5 \dfrac{8}{\sqrt{5}} . This can be seen easily via basic geometrical analytics. (I used the formula of the distance between a point and an iperplane, and the fact that the point (1,1,1,2,3) belongs to the 3rd-dimensional sphere, and the Pythagorean theorem). The max value of x 5 x_5 is attained at the point P P of the 3rd-dimensional sphere that is as far as possible from the iperplane x 5 = 8 / 5 x_5 = 8/5 . We can discover it considering that from the center C C the point P P has distance 8 / 5 8/\sqrt{5} but since P P belongs to the iperplane x 1 + x 2 + x 3 + x 4 + x 5 = 8 x_1+x_2+x_3+x_4+x_5 =8 the length 8 / 5 8/\sqrt{5} must lie on this plane (we cannot travel in the (0,0,0,0,1) direction) so the max distance we can reach from the plane x 5 = 8 / 5 x_5 = 8/5 is the projection of C P CP along the line C + ( 0 , 0 , 0 , 0 , 1 ) t C + (0,0,0,0,1)t line. Due to the special position of the plane we can find this projection searching the side of a 5th-dimensional cube whose main diagonal is r = 8 / 5 r = 8/\sqrt{5} . This is in fact r 5 = 8 5 \dfrac{r}{\sqrt{5}} = \dfrac{8}{5} . So have an estimate of the further distance we can gain (from the centre C C ) along the direction (0,0,0,0,1). Since the 5th coordinate of C C is 8 / 5 8/5 we have the estimate that the 5th coordinate point of the 3rd dimensional sphere (intersection of the 4th dimensional sphere and the iperplane) is lower or equal that 8/5 + 8/5 = 16 5 \dfrac{16}{5} . We must still prove that this value is attained. We plug x 5 x_5 and due to the symmetry of problem x = x 1 = x 2 = x 3 = x 4 x=x_1=x_2=x_3=x_4 and we find 4 x + 16 5 = 8 x = 6 5 4x + \dfrac{16}{5} = 8 \ \Rightarrow x = \dfrac{6}{5} and we check that this also satisfies the 4th dimensional sphere equation since 4 36 25 + 256 25 = 400 25 = 16 4 \cdot \dfrac{36}{25} + \dfrac{256}{25} = \dfrac{400}{25} = 16

Andrea Palma - 1 year, 5 months ago
Sanjeet Raria
Sep 9, 2014

Using Cauchy-Schwarz:

( i = 1 n x i n ) 2 i = 1 n ( x i ) 2 n \left(\frac{\sum_{i=1}^nx_{i}}{n}\right)^2\le\frac{\sum_{i=1}^n(x_{i})^2}{n} We have, ( x 1 + x 2 + + x 4 4 ) 2 x 1 2 + x 2 2 + . . + x 4 2 4 \left(\frac{x_1+x_2+…+x_4}{4}\right)^2\le\frac{x_1^2+x_2^2+..+x_4^2}{4} ( 8 x 5 4 ) 2 16 x 5 2 4 \left(\frac{8-x_5}{4}\right)^2\le \frac{16-x_5^2}{4} On solving, x 5 ( x 5 16 / 5 ) 0 x_5(x_5-16/5)\le 0 Hence maximum value of x 5 = 16 5 . x_5=\boxed{\frac{16}{5}}.

Could you please tell me what is the error in following solution.

as sum of square is 16, so highest possible value of x5^{2} will be 16 hence x5=4 =4/1

=> y1=4, y2=1 => so answer = 9

Chandan Baranwal - 6 years, 9 months ago

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This implies x1=x2=x3=x4=0 from equation 2. Equation 1 will not be satisfied any more. Hence x5 can not be equal to 4.

mohit chawla - 6 years, 9 months ago
Lawrence Bush
Sep 8, 2014

We are given that:

x 1 + x 2 + x 3 + x 4 + x 5 = 8 x 1 + x 2 + x 3 + x 4 = 8 x 5 x_1+x_2+x_3+x_4+x_5=8 \Rightarrow x_1+x_2+x_3+x_4=8-x_5 .

( x 1 + x 2 + x 3 + x 4 ) 2 = ( 8 x 5 ) 2 (x_1+x_2+x_3+x_4)^{2}=(8-x_5)^{2} .

x 1 2 + 2 x 1 x 2 + x 2 2 + x 3 2 + 2 x 3 x 4 + x 4 2 + 2 ( x 1 + x 2 ) ( x 3 + x 4 ) = 64 16 x 5 + x 5 2 x_1^{2}+2x_1x_2+x_2^{2}+x_3^{2}+2x_3x_4+x_4^{2}+2(x_1+x_2) (x_3+x_4)=64-16x_5+x_5^{2} ( 1 ) (1)

We are also given that:

x 1 2 + x 2 2 + x 3 2 + x 4 2 = 16 x 5 2 x_1^{2}+x_2^{2}+x_3^{2}+x_4^{2}=16-x_5^{2} . ( 2 ) (2) .

Notice that:

( 1 ) ( 2 ) x 5 2 8 x 5 + 24 = x 1 x 2 + x 3 x 4 + ( x 1 + x 2 ) ( x 3 + x 4 ) (1)-(2) \Rightarrow x_5^{2}-8x_5+24=x_1x_2+x_3x_4+(x_1+x_2)(x_3+x_4) . ( 3 ) (3)

To find the maximum of x 5 x_5 we must first find the maximum of the R H S RHS ,or:

x 1 x 2 + x 3 x 4 + ( x 1 + x 2 ) ( x 3 + x 4 ) = x 1 x 2 + x 3 x 4 + x 1 x 3 + x 1 x 4 + x 2 x 3 + x 2 x 4 x_1x_2+x_3x_4+(x_1+x_2)(x_3+x_4)= x_1x_2+x_3x_4+x_1x_3+x_1x_4+x_2x_3+x_2x_4 .

We know that:

x 1 x 2 x 1 2 + x 2 2 2 x_1x_2 \leq \frac{x_1^{2}+x_2^{2}}{2}

x 3 x 4 x 3 2 + x 4 2 2 x_3x_4 \leq \frac{x_3^{2}+x_4^{2}}{2} .

x 1 x 3 x 1 2 + x 3 2 2 x_1x_3 \leq \frac{x_1^{2}+x_3^{2}}{2}

x 1 x 4 x 1 2 + x 3 2 2 x_1x_4 \leq \frac{x_1^{2}+x_3^{2}}{2}

x 2 x 3 x 2 2 + x 3 2 2 x_2x_3 \leq \frac{x_2^{2}+x_3^{2}}{2}

x 2 x 4 x 2 2 + x 4 2 2 x_2x_4 \leq \frac{x_2^{2}+x_4^{2}}{2}

Summing these up we get:

x 1 x 2 + x 3 x 4 + x 1 x 3 + x 1 x 4 + x 2 x 3 + x 2 x 4 3 2 ( x 1 2 + x 2 2 + x 3 2 + x 4 2 ) = 24 3 x 5 2 2 x_1x_2+x_3x_4+x_1x_3+x_1x_4+x_2x_3+x_2x_4 \leq \frac{3} {2}*(x_1^{2}+x_2^{2}+x_3^{2}+x_4^{2})=24 -\frac{3x_5^{2}}{2} .

Substituting these final value on the R H S RHS of ( 3 ) (3) ,we get:

x 5 2 8 x 5 + 24 = 24 3 x 5 2 2 x_5^{2}-8x_5+24=24-\frac{3x_5^{2}}{2} .

Multiply both sides by 2 2 and after making some arrangements we get:

5 x 5 2 16 x 5 = 0 x 5 = 16 5 5x_5^{2}-16x_5=0 \Rightarrow \boxed{x_5=\frac{16}{5}} .

Nice solution! I actually thought of doing the same thing, but decided to go ahead with the Lagrange multipliers method. You just have to remember that you need to check that the set of equations you get from all the inequalities is consistent. In your case it was because what you end up with reduces to x 1 = x 2 = x 3 = x 4 x_1=x_2=x_3=x_4 . If you don't know what I'm talking about, try this problem: https://brilliant.org/practice/classical-inequalities-applications-level-4-5/?p=4

James Wilson - 3 years, 5 months ago

Could have been shorter.

Rudrayan Kundu - 2 years, 7 months ago
Daniel Fadul
Sep 18, 2014

This problem is solvable because in reality is a system of two equations and two variables. This can be seen by noticing the fact that x 1 = x 2 = x 3 = x 4 0 x_1=x_2=x_3=x_4\neq0 . There are two conditions implicit in the latter equation: the first, all four variables being different from zero ensures that both equation are satisfied; the second, if all variables are equal x 5 x_5 can have a maximum value (give it a thought).

Given the latter, we have now the following:

4 x 1 + x 5 = 8 4x_1+x_5=8

4 x 1 2 + x 5 2 = 16 4x_1^2+x_5^2=16

Solving for x 5 x_5 one gets that it must be either zero or 16 5 \frac{16}{5} , and the final answer follows.

Moreover, if one writes the original equations as

i = 1 n x i = 8 \sum_{i=1}^n x_i=8

i = 1 n x i 2 = 16 \sum_{i=1}^n x_i^2=16 ,

the problem is solvable if n 4 n\geq4 ; nevertheless, if n = 4 n=4 , x 1 = x 2 = x 3 = x 4 = 2 x_1=x_2=x_3=x_4=2 - in other words, there is no "maximum variable". For n 5 n\geq5 we will have a maximum value for x 5 x_5 (give it another thought!)

Exactly the same solution I have.

Yes, it is quite obvious that x 1 = x 2 = x 3 = x 4 x_1=x_2=x_3=x_4 for x 5 x_5 to be at its maximum. Any increase in any of those would mean a decrease in x 5 x_5 . Any decrease would mean relatively increasing the others resulting to a decreased x 5 x_5 .

Roman Frago - 6 years, 3 months ago
Reynan Henry
Jan 18, 2017

i = 1 5 x i 2 = 2 i = 1 5 x i 5 = i = 1 5 ( x i 1 ) 2 ( x 5 1 ) 2 x 5 1 + 5 \sum_{i=1}^5 x_i^2= 2\sum_{i=1}^5x_i\implies 5=\sum_{i=1}^5 (x_i-1)^2\ge (x_5-1)^2\implies x_5\le 1+\sqrt{5}

Solved also by finding cross section between 5-dimensional plane and 5-dimensional sphere.

Cantdo Math
Apr 9, 2020

we see that, x 1 + x 2 + x 3 + x 4 = 8 x 5 x_1+x_2+x_3+x_4=8-x_5 and x 1 2 + . . x 4 2 = 16 x 5 2 x_1^2+..x_4^2=16-x_5^2 .Then using QM-AM inequality suffices.

James Wilson
Dec 17, 2017

Joseph-Louis Lagrange Multipliers should be one of your many friends! Let f ( x 1 , x 2 , x 3 , x 4 ) = x 5 = 16 x 1 2 x 2 2 x 3 2 x 4 2 f(x_1,x_2,x_3,x_4)=x_5=\sqrt{16-x_1^2-x_2^2-x_3^2-x_4^2} be the objective function to be maximized. Then let the constraint be g ( x 1 , x 2 , x 3 , x 4 ) = x 1 + x 2 + x 3 + x 4 + 16 x 1 2 x 2 2 x 3 2 x 4 2 = 8 g(x_1,x_2,x_3,x_4)=x_1+x_2+x_3+x_4+\sqrt{16-x_1^2-x_2^2-x_3^2-x_4^2}=8 . Then f x k = x k 16 x 1 2 x 2 2 x 3 2 x 4 2 = x k x 1 + x 2 + x 3 + x 4 8 \frac{\partial f}{\partial x_k}=\frac{-x_k}{\sqrt{16-x_1^2-x_2^2-x_3^2-x_4^2}}=\frac{x_k}{x_1+x_2+x_3+x_4-8} for k = 1 , 2 , 3 , 4 k=1,2,3,4 , and g x k = 1 + x k 16 x 1 2 x 2 2 x 3 2 x 4 2 = 1 + x k x 1 + x 2 + x 3 + x 4 8 \frac{\partial g}{\partial x_k}=1+\frac{-x_k}{\sqrt{16-x_1^2-x_2^2-x_3^2-x_4^2}}=1+\frac{x_k}{x_1+x_2+x_3+x_4-8} for k = 1 , 2 , 3 , 4 k=1,2,3,4 . This leads to the linear system: ( 2 1 λ ) x k + m k x m = 8 \Big(2-\frac{1}{\lambda}\Big)x_k+\sum_{m\neq k}x_m=8 . Subtracting two of the equations gives ( 1 1 λ ) x k = ( 1 1 λ ) x j \Big(1-\frac{1}{\lambda}\Big)x_k=\Big(1-\frac{1}{\lambda}\Big)x_j for i j i\neq j . If one assumes λ = 1 \lambda=1 , then x 1 + x 2 + x 3 + x 4 = 8 x_1+x_2+x_3+x_4=8 , which leads to x 5 = 0 x_5=0 (a minimum). Assuming λ 1 \lambda\neq 1 leads to x 1 = x 2 = x 3 = x 4 x_1=x_2=x_3=x_4 , which, in turn, leads to the solution x 5 = 16 5 x_5=\frac{16}{5} .

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