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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x

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Manager  Joined: 09 Feb 2013
Posts: 111
If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 52% (01:44) correct 48% (01:48) wrong based on 561 sessions

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If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is

A. Less than 10
B. Greater than or equal to 10 and less than 14
C. Greater than 14 and less than 19
D. Greater than 19 and less than 23
E. Greater than 23

Originally posted by emmak on 11 Feb 2013, 10:56.
Last edited by Bunuel on 11 Feb 2013, 14:30, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 60467
Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x  [#permalink]

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11
20
emmak wrote:
If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is

A. Less than 10
B. Greater than or equal to 10 and less than 14
C. Greater than 14 and less than 19
D. Greater than 19 and less than 23
E. Greater than 23

$$x^2 + y^2 = 100$$ --> $$(x+y)^2-2xy=100$$ --> $$(x+y)^2=100+2xy$$ --> $$x+y=\sqrt{100+2xy}$$.

We need to maximize $$x+y$$, thus we need to maximize $$\sqrt{100+2xy}$$. To maximize $$\sqrt{100+2xy}$$ we need to maximize $$xy$$.

Now, for given sum of two numbers, their product is maximized when they are equal, hence from $$x^2 + y^2 = 100$$ we'll have that $$x^2y^2$$ (or which is the same xy) is maximized when $$x^2=y^2$$.

In this case $$x^2 + x^2 = 100$$ --> $$x=\sqrt{50}$$.

So, we have that $$\sqrt{100+2xy}$$ is maximized when $$x=y=\sqrt{50}$$, so the maximum value of $$\sqrt{100+2xy}$$ is $$\sqrt{100+2*\sqrt{50}*\sqrt{50}}=\sqrt{200}\approx{14.1}$$.

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Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x  [#permalink]

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9
1
I tried a more crude method:

$$x^2 + y^2 = 100$$

We know $$6^2 + 8^2$$ is one of the options which will lead to x+y = 14. So options A & B are out.

We also know $$7^2 + 7^2$$ we get just $$98<100$$. So something slightly more than $$7$$ i.e. $$7.1$$ or whatever that will lead to the answer of $$100$$. So the maximum value of $$x+y$$ is just over $$14$$. Any other combination of x & y cannot be more than this value of just over 14. So answer is C.
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Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x  [#permalink]

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7
An elegant solution is looking at this problem geometrically.

x^2 + y^2 = 100 is a circumference with radius 10 and center in (0,0).

The equation x + y = k (where we want to maximize k) is a set of infinite parallel negative slope (-45 deg) lines.

Now note that k is the Y INTERSECT of all such lines. In order to maximize this y intersect (and therefore maximize k) we need to find the line of the set that is tangent to the circunference in the 1st quadrant.

Now this becomes a (fairly easy) geometry problem.

Drawing it out you'll have no problem seeing that
Y intersect = k = RADIUS / Sin(45 deg) = 10.sqrt(2) = approx 14,1

Posted from my mobile device
Intern  Joined: 23 Feb 2013
Posts: 8
Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x  [#permalink]

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Dear Bunuel,

Can you please explain this statement,"for given sum of two numbers, their product is maximized when they are equal". Can you give some theory and examples for this statement to be true ? I would be happy to understand this. Thank you.
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Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x  [#permalink]

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11
6
emmak wrote:
If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is

A. Less than 10
B. Greater than or equal to 10 and less than 14
C. Greater than 14 and less than 19
D. Greater than 19 and less than 23
E. Greater than 23

You can use the logical approach to get the answer within seconds.

First you need to understand how squares work - As you go to higher numbers, the squares rise exponentially (obviously since they are squares!)
What I mean is
2^2 = 4
3^2 = 9
4^2 = 16
(as we increase the number by 1, the square increases by more than the previous increase. From 2^2 to 3^2, the increase is 9-4= 5 but from 3^2 to 4^2, the increase is 16 - 9 = 7... As we go to higher numbers, the squares will keep increasing more and more.)

If we want to keep the square at 100 but maximize the sum of the numbers, we should try and make the numbers as small as possible so that their contribution in the square doesn't make the other number very small i.e. if you take one number almost 10, the other number will become very very small and the sum will not be maximized. If one number is made small, the other will become large, hence both numbers should be equal to maximize the sum.
So the square of each number should 50 i.e. each number should be a little more than 7.
Both numbers together will give us something more than 14.

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Posts: 60467
Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x  [#permalink]

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2
Backbencher wrote:
Dear Bunuel,

Can you please explain this statement,"for given sum of two numbers, their product is maximized when they are equal". Can you give some theory and examples for this statement to be true ? I would be happy to understand this. Thank you.

If a+c=k, then the product ab is maximized when a=b.

For example, if given that a+b=10, then ab is maximized when a=b=5 --> ab=25.

Hope it's clear.
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Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x  [#permalink]

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VeritasPrepKarishma wrote:
emmak wrote:
If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is

A. Less than 10
B. Greater than or equal to 10 and less than 14
C. Greater than 14 and less than 19
D. Greater than 19 and less than 23
E. Greater than 23

You can use the logical approach to get the answer within seconds.

First you need to understand how squares work - As you go to higher numbers, the squares rise exponentially (obviously since they are squares!)
What I mean is
2^2 = 4
3^2 = 9
4^2 = 16
(as we increase the number by 1, the square increases by more than the previous increase. From 2^2 to 3^2, the increase is 9-4= 5 but from 3^2 to 4^2, the increase is 16 - 9 = 7... As we go to higher numbers, the squares will keep increasing more and more.)

If we want to keep the square at 100 but maximize the sum of the numbers, we should try and make the numbers as small as possible so that their contribution in the square doesn't make the other number very small i.e. if you take one number almost 10, the other number will become very very small and the sum will not be maximized. If one number is made small, the other will become large, hence both numbers should be equal to maximize the sum.
So the square of each number should 50 i.e. each number should be a little more than 7.
Both numbers together will give us something more than 14.

Sorry to open this post after so long time again but i got a bit confused reg the solution.
I agree with the solution that maximizing sum is by making both the numbers x and y equal resulting in value greater than 14.
but what about checking whether the value is less than 19 or not.
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Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x  [#permalink]

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Mechmeera wrote:
VeritasPrepKarishma wrote:
emmak wrote:
If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is

A. Less than 10
B. Greater than or equal to 10 and less than 14
C. Greater than 14 and less than 19
D. Greater than 19 and less than 23
E. Greater than 23

You can use the logical approach to get the answer within seconds.

First you need to understand how squares work - As you go to higher numbers, the squares rise exponentially (obviously since they are squares!)
What I mean is
2^2 = 4
3^2 = 9
4^2 = 16
(as we increase the number by 1, the square increases by more than the previous increase. From 2^2 to 3^2, the increase is 9-4= 5 but from 3^2 to 4^2, the increase is 16 - 9 = 7... As we go to higher numbers, the squares will keep increasing more and more.)

If we want to keep the square at 100 but maximize the sum of the numbers, we should try and make the numbers as small as possible so that their contribution in the square doesn't make the other number very small i.e. if you take one number almost 10, the other number will become very very small and the sum will not be maximized. If one number is made small, the other will become large, hence both numbers should be equal to maximize the sum.
So the square of each number should 50 i.e. each number should be a little more than 7.
Both numbers together will give us something more than 14.

Sorry to open this post after so long time again but i got a bit confused reg the solution.
I agree with the solution that maximizing sum is by making both the numbers x and y equal resulting in value greater than 14.
but what about checking whether the value is less than 19 or not.

Hi Mechmeera,

The maximum value has been calculated when x = y = $$5\sqrt{2}$$

i.e. x+y will have maximum value of $$5\sqrt{2}$$ + $$5\sqrt{2}$$ = $$10\sqrt{2}$$ = 14.14

Since the value of x+y can't exceed beyond the calculated value therefore

we can conclude that it is less that 19 and greater than 14.
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Re: Problem Solving: Number Properties  [#permalink]

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BetaRayBryan wrote:
If $$x^2 + y^2 = 100$$, $$x\geq{0}$$ and $$y\geq{0}$$, the maximum value of x + y must be which of the following?

A) Less than 10

B) Greater than or equal to 10 and less than 14

C) Greater than 14 and less than 19

D) Greater than 19 and less than 23

E) Greater than 23

Would someone be able to walk me through this? I'm sort of unclear on the reasoning behind the correct answer.. Thank you so much!

Hi

$$x^2+y^2=100=10^2$$..
Least- when the number are far away so 10 and 0, x+y=10
Max- when X and y are close by.. 7^2=49 so both as 7 gives us 7^2+7^2=98
And x+y =7+7=14, so ans will be slightly MORE than 14
C
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Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x  [#permalink]

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Thank you all, I got the same answer but with bit of confused, I AGREED with answer 14.1 but the answer also given as > 14 to <19. So, it implies that the answer could be anywhere from 14.1 to 18.9? I understand that 14.1 is the possible answer, then why we have answer given <19?

Thank you.
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GMAT 1: 700 Q49 V36 If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x  [#permalink]

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caioguima wrote:
An elegant solution is looking at this problem geometrically.

x^2 + y^2 = 100 is a circumference with radius 10 and center in (0,0).

The equation x + y = k (where we want to maximize k) is a set of infinite parallel negative slope (-45 deg) lines.

Now note that k is the Y INTERSECT of all such lines. In order to maximize this y intersect (and therefore maximize k) we need to find the line of the set that is tangent to the circunference in the 1st quadrant.

Now this becomes a (fairly easy) geometry problem.

Drawing it out you'll have no problem seeing that
Y intersect = k = RADIUS / Sin(45 deg) = 10.sqrt(2) = approx 14,1

Posted from my mobile device

chetan2u
Bunuel
Could you explain this solution in further detail please. Thanks.
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Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x  [#permalink]

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emmak wrote:
If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is

A. Less than 10
B. Greater than or equal to 10 and less than 14
C. Greater than 14 and less than 19
D. Greater than 19 and less than 23
E. Greater than 23

we can use the formula of a^2=b^2+c^2
knowing 10^2 = 6^2 + 8^2
so max value would be 8+6 ; 14
option C Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x   [#permalink] 22 Feb 2019, 10:02
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