If (x^2) + (y^2)=100, x>=0, and y>=0, the maximum value of x+y must be which of the following?
A. Less than 10
B. Greater or equal to 10 and less than 14
C. Greater than 14 and less than 19
D. Greater than 19 and less that 23
E. Greater than 23
*Can anyone explain this in a simple way?
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Tough Algebra Number Properties Question
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tohellandback
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is it C?
if it is, then I will post the explanation.
x^2+y^2=100
(x+y)^2-2xy=100
(x+y)^2=100+2xy
if we can find the max value of xy, we are done
a number property:
a+b=x
ab will be max when a=b=x/2
max value of x^2*Y^2=50*50
max value of xy=50
max val of (x+y)^2=100+2*50=200
max val of x+y=sqrt(200)
answer C
if it is, then I will post the explanation.
x^2+y^2=100
(x+y)^2-2xy=100
(x+y)^2=100+2xy
if we can find the max value of xy, we are done
a number property:
a+b=x
ab will be max when a=b=x/2
max value of x^2*Y^2=50*50
max value of xy=50
max val of (x+y)^2=100+2*50=200
max val of x+y=sqrt(200)
answer C
Last edited by tohellandback on Mon Aug 10, 2009 6:58 pm, edited 1 time in total.
The powers of two are bloody impolite!!
Hmm. Obviously x and y can be non integers since nothing has been specified. So I just tried a bunch of combinations and found that C fits because x^2 + y^2 can be written as 50 + 50 which is (7.something)^2 + (7.something)^2 and so x = 7.smthng and y is 7.smthng and their sum is 14.smthng which is greater than 14 however no combination will yield values greater than 19 so I chose C.
200 or 800. It don't matter no more.
