Suppose that |x + y| + |x -y| = 2. What is the maximum possible value

Question

Suppose that |x + y| + |x -y| = 2. What is the maximum possible value of x^2 - 6x + y^2?

A. 5
B. 6
C. 7
D. 8
E. 9

eswarchethu135 · Accepted Answer

if we open mod there are 4 cases possible

Case 1) Both are positive

x + y + x - y = 2
2x = 2
x = 1

Case 2) |x + y| is negative

-x - y + x - y = 2
-2y = 2
y = -1

Case 3) |x - y| is negative
x + y - x + y = 2
2y = 2
y = 1

Case 4) Both are negative
- x - y - x + y = 2
-2x = 2
x = -1

So from the above 4 cases, we can say  x = \pm{1}  and  y = \pm{1}

Substituting in the eq:  x^2 - 6x + y^2

1 - 6(\pm{1}) + 1

This value will be maximum when x = -1

= 1 -6(-1) + 1

=8

OPTION:  D

chetan2u · Answer

We can find our solution by logic.
(I) Both the terms |x + y| and |x -y| are at least 0, and cannot be more than 2 as their sum itself is given as 2.
(II) If you have to maximize  x^2 - 6x + y^2 , we will try to maximize -6x as that will be the largest value, so x should be negative.

Now |x| and |y| should not be greater than 1, so both can be 1..
x has to be negative, so x=-1, and y =1..
|x + y| + |x -y| =|-1+1|+|-1-1|= 2...Yes

Value of  x^2 - 6x + y^2  will become  (-1)^2 - 6(-1) + (-1)^2=1+6+1=8

D

gvij2017 · Answer

How I approached this question is as follows. 
I know that every term in equation is positive. Therefore, I squared both sides. 
And got (x^2 + y^2)= 2.

Now I have maximize the x^2 + y^2 - 6x
To maximize 2-6x, x must be negative and can be -1. 
So, maximum value is 8.

I think my approach to such questions in which both sides are positive is correct.

foryearss · Answer

Hi , I'm curious , when you square both sides , you got 2 *|x + y| * |x -y| in your way , how you did deal with that?

vinayakvaish · Answer

(|x+y| + |x-y|) ^2 = 4
2(x^2 + y^2) + 2( x^2-y^2)=4
4x^2=4
x=+/- 1

to maximize, x must be -1

put this in original equation, y can be 0 or 1

since we are looking to maximize, choose 1

Answer= 1+6+1= 8

drasticgre · Answer

Hei vinayakvaish
your 2nd line - isnt it possible 2(x^2 + y^2)  - 2( x^2-y^2) =4 ?

hadimadi · Answer

Hi,

drasticgre ,  vinayakvaish

Let a=root((x+y)^2), b=root((x-y)^2)

We have:

a+b=2, now we square to get
(a+b)^2=4
a^2+2ab+b^2=4, plugging in our initial terms we get
x^2+2xy+y^2+2(lx+yl*lx-yl)+x^2-2xy+y^2=4
2x^2+2y^2+2*lx^2-y^2l=4
x^2+y^2+lx^2-y^2l=2

C1, lxl>lyl

2x^2=2
-> x_1,2=1, -1

x has to be the negative solution, because the term x^2-6x will then be greatest. The corresponding y is then -1 or 1, and we get 8
for our equation for both y's

C2, lxl<lyl

2y^2=2
-> y_1,2=1, -1

We again need a negative x, so in total, we have 8 again

Bunuel ,  nick1816 , could you kindly check if this solution is correct? I feel like something is wrong or it's too lengthy

Oppenheimer1945 · Answer

Max((x-3)^2+y^2-9) = max dist of (3,0) from a point in square=4^2+1^2-9=8

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Suppose that |x + y| + |x -y| = 2. What is the maximum possible value

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Suppose that |x + y| + |x -y| = 2. What is the maximum possible value

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