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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
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
emmak
\(x^2 + y^2 = 100\) --> \((x+y)^2-2xy=100\) --> \((x+y)^2=100+2xy\) --> \(x+y=\sqrt{100+2xy}\).
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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emmak
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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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.
\(x^2 + y^2 = 100\)
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.
General Discussion
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
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
Posts: 8
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.
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.
Backbencher
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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VeritasPrepKarishma
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.
Can you please explain.
Mechmeera
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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BetaRayBryan
Hi
I will be glad to help you, but please follow the rules for starting a thread.
\(x^2+y^2=100=10^2\)..
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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emmak
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
Posts: 1,015
caioguima
Nice ...I would like to extend it using calculus: Might be helpful for people good at calculus:-
Soln:
x^2 + y^2 = 100 is the equation of a circle with a radius of 10.
x and y can be written as r.sin(\(\theta\)) and r.cos(\(\theta\)) and we want to mazimize x+y Since r is constant 10, we only need to differentiate by \(\theta\), 10 [sin(\(\theta\)) + cos(\(\theta\))] which is our objective function.
Find first order condition which is cos(\(\theta\))-sin(\(\theta\)) and set the equation to zero to get sin(\(\theta\)) = cos(\(\theta\))
Which implies \(\theta\) = 45 degrees
or r. Sin(45 deg) + r Cos(45 deg) = 10 [1/sqrt(2)+1/sqrt(2)] = approx 14.1
Posts: 3
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
In this question we know x,y is >=O so it can be integer/ fraction.
Now given x^2+y^2 = 100
We can wright this (x+y)^2 = 100+ 2xy
Seeing this we can eliminate option A. As it will never be less than 10.
Similarly we can eliminate E&D. As it can't be greater than or equal to 23 in any case Assuming x,y is an integer so best possible value would be 6,8 . So 100+2*6*8 = 196
And sq.root of 196 is 14 so our ans should be around 14. Since there can be decimals value of x,y possible that can make x^2+y^2=100 so it can't be less than 14 atlist. So we can eliminate option B.
And we are left with C. Which will be our best possible ans.
Posted from my mobile device
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
In this question we know x,y is >=O so it can be integer/ fraction.
Now given x^2+y^2 = 100
We can wright this (x+y)^2 = 100+ 2xy
Seeing this we can eliminate option A. As it will never be less than 10.
Similarly we can eliminate E&D. As it can't be greater than or equal to 23 in any case Assuming x,y is an integer so best possible value would be 6,8 . So 100+2*6*8 = 196
And sq.root of 196 is 14 so our ans should be around 14. Since there can be decimals value of x,y possible that can make x^2+y^2=100 so it can't be less than 14 atlist. So we can eliminate option B.
And we are left with C. Which will be our best possible ans.
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Asked: If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is
Maximum value of x+y is when x=y
x^2 + y^2 = 100
2x^2 = 100
x^2 = 50 = 2*25
\(x = y = 5\sqrt{2}\)
\(x + y = 10\sqrt{2} = 14.14\)
IMO C
Maximum value of x+y is when x=y
x^2 + y^2 = 100
2x^2 = 100
x^2 = 50 = 2*25
\(x = y = 5\sqrt{2}\)
\(x + y = 10\sqrt{2} = 14.14\)
IMO C
Posts: 1
if X is 8 and Y is 6 = 100; 8+6= 14. But how does exactly 14 fit in to the answers?
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Kinshook
Major Poster
Joined: 03 Jun 2019
Last visit: 10 Sep 2026
Posts: 6,128
Location: India
Schools: HBS '28 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
emmak
If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is
The maximum value of x+y is when x=y
\(xˆ2 + xˆ2 = 100; xˆ2 = 50; x=5\sqrt{2}\)
\(x + y = 2x = 10\sqrt{2} = 10*1.414 = 14.14 \) approx
IMO C
If x^2 + y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is
The maximum value of x+y is when x=y
\(xˆ2 + xˆ2 = 100; xˆ2 = 50; x=5\sqrt{2}\)
\(x + y = 2x = 10\sqrt{2} = 10*1.414 = 14.14 \) approx
IMO C
Answer may be correct but the approach is incorrect.
The max value of x + y does not always occur at x=y. Take x^2 + 2y^2 = 1 for instance. The max occurs at x = 2y. This questions is perfectly symmetric so that works out.
In general, max/min always occurs at dy/dx=0. Simply differentiating x + y gives at what x the answer is max/min.
Just apply d(x^n)/dx=n*x^n−1 and put it back in the equation.
The max value of x + y does not always occur at x=y. Take x^2 + 2y^2 = 1 for instance. The max occurs at x = 2y. This questions is perfectly symmetric so that works out.
In general, max/min always occurs at dy/dx=0. Simply differentiating x + y gives at what x the answer is max/min.
Just apply d(x^n)/dx=n*x^n−1 and put it back in the equation.
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