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

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If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink] New post 11 Feb 2013, 10:56
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 11 Feb 2013, 14:30, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink] New post 11 Feb 2013, 14:44
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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


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}.

Answer: C.
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Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink] New post 11 Feb 2013, 15:05
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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] New post 13 Feb 2013, 09:31
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

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Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink] New post 11 Mar 2013, 18:17
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] New post 11 Mar 2013, 20:43
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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


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.

Answer (C)
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Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x [#permalink] New post 12 Mar 2013, 06:50
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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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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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Re: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x   [#permalink] 12 Mar 2013, 06:50
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