rjain84 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

OA: C
Source: Veritas Prep

for inequalities question just remember this small trick," To maximize the sum, no.s should be as far as possible from each other and to maximize the multiplication, no.s should be as close as possible."

Now lets take your question. We need to maximize the sum so x and y should be as far as possible

0^2 + 10^2 = 100

x+y = 10

answer is b.


If you still have confusion take any no. say 16, (l*b)area of rectangle. Now you want to find the max possible perimeter of this rectangle given length and breadth both are integers.

possible combinations are (1,16), (2,8), (4,4). Now (1,16) will give you the max perimeter of 17.

Hope this helps.

Cheers,
Hemant

hemant_rajput wrote:

rjain84 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

OA: C
Source: Veritas Prep

for inequalities question just remember this small trick," To maximize the sum, no.s should be as far as possible from each other and to maximize the multiplication, no.s should be as close as possible."

Now lets take your question. We need to maximize the sum so x and y should be as far as possible

0^2 + 10^2 = 100

x+y = 10

answer is b.


If you still have confusion take any no. say 16, (l*b)area of rectangle. Now you want to find the max possible perimeter of this rectangle given length and breadth both are integers.

possible combinations are (1,16), (2,8), (4,4). Now (1,16) will give you the max perimeter of 17.

Hope this helps.

Cheers,
Hemant

Thanks for the explanation but the OA is C not B.

Anurag@Gurome wrote:

rjain84 wrote:If x^2+y^2 = 100, and x≥0, and y≥0, the maximum value of x + y is

x² + y² = 100
Hence, 0 ≤ x ≤ 10 and 0≤ y ≤ 10
Some obvious possible values for x and y are {x = 10, y = 0} or {x = 6, y = 8}
In first case, (x + y) = 10 ---> Discard option 1
In second case, (x + y) = 14 ---> Discard option 2

Now, 7² = 49 ---> 7² + 7² = 98 is just less than 100
Hence, if we take x = y = something just more than 7 such that (x² + y²) = 100, we will get the maximum value of (x + y) which will be just more than 14.

The correct answer is C.

Wonderful explanation! Thanks a lot Anurag!

Anurag@Gurome wrote:Algebraic Method:
(x + y)² = (x² + y² + 2xy) = (100 + 2xy) ---> (x + y) = √(100 + 2xy)

To maximize (x + y), we need to maximize xy which we will do by maximizing x²y².
Now, remember this rule if you don't know it yet,

• Given the sum of two numbers, their product will be maximum when they are equal to each other.

Now, we know x² + y² = 100.
Hence, x²y² will be maximum when x² = y² = 100/2 = 50
So, maximum value of x²y² = 50*50
So, maximum value of xy = √(50*50) = 50
So, maximum value of (x + y) = √(100 + 2*50) = √200 = 10√2 = 10*1.414 = 14.14

The correct answer is C.

Thanks for the tip. Can we also say the same for the given product? Given the product of two numbers, their sum will be maximum when both are equal?

Anurag@Gurome wrote:

rjain84 wrote:Can we also say the same for the given product? Given the product of two numbers, their sum will be maximum when both are equal?

No. The reverse is not true.
To illustrate with a simple example, take xy = 4 ---> (4 + 1) > (2 + 2)

Gotcha! Thank you Anurag

oops! I messed up the concepts.
@Anurag, thanks for correcting my mistake.

Kudos,
Hemant