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If x and y are integers such that (x+1)^2 less than equal to [#permalink]
25 Jul 2010, 17:51

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If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy.

Re: In equalities how to handle an expression with squares [#permalink]
25 Jul 2010, 17:58

gmatrant wrote:

If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy.

In equalities how to handle an expression which is squared Does the above equation (x+1)^2 <= 36 mean |x+1| < (+6 or -6)

I then get 4 equations.. and I am am not able to proceed. Can you someone please explain how such questions are to be handled.

Re: In equalities how to handle an expression with squares [#permalink]
12 Aug 2010, 12:01

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gmatrant wrote:

If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy.

In equalities how to handle an expression which is squared Does the above equation (x+1)^2 <= 36 mean |x+1| < (+6 or -6)

I then get 4 equations.. and I am am not able to proceed. Can you someone please explain how such questions are to be handled.

If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy.

(y-1)^2<{64} --> {-\sqrt{64}}<{y-1}<{\sqrt{64}} --> {-8}<{y-1}<{8} --> {-7}<{y}<{9}, as y is an integer we can rewrite this inequality as {-6}\leq{y}\leq{8}.

We should try extreme values of x and y to obtain min and max values of xy:

Min possible value of xy is for x=-7 and y=8 --> xy=-56; Max possible value of xy is for x=-7 and y=-6 --> xy=42.

Solving with absolute values gives the same results:

(x+1)^2\leq{36} means |x+1|\leq{6} --> {-7}\leq{x}\leq{5}. (y-1)^2<{64} means |y-1|<{8} --> {-7}<{y}<{9}.

Re: In equalities how to handle an expression with squares [#permalink]
28 Aug 2012, 03:10

1

This post received KUDOS

Expert's post

pallavisatsangi wrote:

Hi Bunuel, " as y is an integer we can rewrite this inequality as -6<= y<=8 ."

I didn't understand as to how can we change the range of y from -7,9 to -6,8 ?

We are not changing the range here.

We have {-7}<{y}<{9}. Now, since y is an integer, then it can take integer values from -6 to 8, inclusive: -6, -5, ..., 6, 7, 8, which can be written as {-6}\leq{y}\leq{8}.

Now solving by the graphical approach, range is between (and inclusive of) -7 and 5.

Similarly solving for the other equation we get the roots in the range of (exclusive of) -7 and 9. Therefore maximum and minimum value of xy can be derived as others before me have explained.

Re: If x and y are integers such that (x+1)^2 less than equal to [#permalink]
05 Dec 2012, 04:06

One thing I notice is that you have to be careful with the exclusivity and inclusivity of ranges. In these questions, you will get it wrong if you thought y = 9 is included in the range.

|x+1| <= 6 This means x is within the range of [-1-6,-1+6] = [-7,5]. This is inclusive of -7 and 5.

|y-1|<8 This means y is within the range of (1-8,1+8) = (-7,9). This is exclusive of -7 and 9.

Now to get the extreme values.

Max value or Positive outcomes: -7 * -6 = 42 or 5 * 8 = 40 ==> 42 wins! Min value of Negative outcomes: -7 * 8 = -56 or -6 * 5 = -30 ==> -56 wins!

Re: If x and y are integers such that (x+1)^2 less than equal to [#permalink]
22 Jul 2013, 08:37

2

This post received KUDOS

gmatrant wrote:

If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy.

Re: If x and y are integers such that (x+1)^2 less than equal to [#permalink]
29 Jul 2013, 01:12

2. If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy. (x+1)^2 < 36 (x+1) < +6 (applying square root on both sides) X < +6-1 X < 5

X >-6-1 X > -7 Range of possible values for x are between -7 and +5

(y-1)^2 < 64 (y-1) <+ 8 (applying square root on both sides) y < +8+1 y < +9

y < -8+1 y > -7 As y is an integer y < 8 or y > -6 Range of possible values for y are between -6 and +8

Range of possible values for xy at their respective highest and lowest levels: -30, 40, 42, -56 Highest value: 42 and lowest value: -56

Re: If x and y are integers such that (x+1)^2 less than equal to [#permalink]
16 Sep 2014, 19:58

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Re: If x and y are integers such that (x+1)^2 less than equal to [#permalink]
05 Dec 2014, 20:37

Bunuel wrote:

pallavisatsangi wrote:

Hi Bunuel, " as y is an integer we can rewrite this inequality as -6<= y<=8 ."

I didn't understand as to how can we change the range of y from -7,9 to -6,8 ?

We are not changing the range here.

We have {-7}<{y}<{9}. Now, since y is an integer, then it can take integer values from -6 to 8, inclusive: -6, -5, ..., 6, 7, 8, which can be written as {-6}\leq{y}\leq{8}.

Hope it's clear.

if it is asking for maximum and minimum value. why did you re-wrote it?

Re: If x and y are integers such that (x+1)^2 less than equal to [#permalink]
06 Dec 2014, 04:54

Expert's post

ikishan wrote:

Bunuel wrote:

pallavisatsangi wrote:

Hi Bunuel, " as y is an integer we can rewrite this inequality as -6<= y<=8 ."

I didn't understand as to how can we change the range of y from -7,9 to -6,8 ?

We are not changing the range here.

We have {-7}<{y}<{9}. Now, since y is an integer, then it can take integer values from -6 to 8, inclusive: -6, -5, ..., 6, 7, 8, which can be written as {-6}\leq{y}\leq{8}.

Hope it's clear.

if it is asking for maximum and minimum value. why did you re-wrote it?

Isn't it explained in the very post you are quoting? _________________

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