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The function f(x,y) is such that f(x,y) = x^2.y^3 & x+y =25. Where x,

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ShalabhsQuants
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The function f(x,y) is such that f(x,y) = x^2.y^3 & x+y =25. Where x, [#permalink] New post   Updated on: 03 Oct 2023, 03:04
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The function f(x,y) is such that f(x,y) = \(x^2\).\(y^3\) & x+y =25. Where x, & y are non-negative integers. Select one minimum possible, & one maximum possible value for f(x,y).
Make only two selections, one in each column.
minimum
f(x,y) 		maximum
f(x,y)
0
29600
337500
400000
infinity
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minimum
f(x,y): 0 maximum
f(x,y): 337500

Originally posted by ShalabhsQuants on 09 Sep 2012, 07:39.
Last edited by BottomJee on 03 Oct 2023, 03:04, edited 6 times in total.
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Re: The function f(x,y) is such that f(x,y) = x^2.y^3 & x+y =25. Where x, [#permalink] New post   09 Sep 2012, 11:36
ShalabhsQuants wrote:
Hello Friends!

Try this question...


I checked the OG13 and QuantReview: they don't use neither the term whole number, nor that of natural number.
I guess, in order to avoid the lack of consensus regarding 0. And in fact, why have so many different definitions?
It is much easier to have just integers, which are positive, negative or zero. Further, you can have non-negative integers if you want to include 0,...

Oh, and I don't think the symbol for infinity should be known on the GMAT.
I would say, not a must to solve such questions for a future test taker.
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Re: The function f(x,y) is such that f(x,y) = x^2.y^3 & x+y =25. Where x, [#permalink] New post   09 Sep 2012, 20:04
Min value of f(x,y) = 0, If we choose either x or y as 0 and the other as 25. The product x^2y^3 becomes 0.
Max value of f(x,y) = Infinity. Since x+y =25, and both are integers, we can choose y as a +ve number such as 10000xxxxxx000 and x as -ve no such as 999999xxxxx75. If we add we will get 25, but if we square x it will be a positive expression and the value of f(x,y) can go up to infinity.
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Re: The function f(x,y) is such that f(x,y) = x^2.y^3 & x+y =25. Where x, [#permalink] New post   09 Sep 2012, 22:25
SOURH7WK wrote:
Min value of f(x,y) = 0, If we choose either x or y as 0 and the other as 25. The product x^2y^3 becomes 0.
Max value of f(x,y) = Infinity. Since x+y =25, and both are integers, we can choose y as a +ve number such as 10000xxxxxx000 and x as -ve no such as 999999xxxxx75. If we add we will get 25, but if we square x it will be a positive expression and the value of f(x,y) can go up to infinity.


As far as Minimum value of f(x,y) is concerned, your answer is correct. But Maximum value is not infinity. Remember x, & y both are non-negative integers. Minimum values of x, & y would be 0 only, whereas maximum as 25.
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Re: The function f(x,y) is such that f(x,y) = x^2.y^3 & x+y =25. Where x, [#permalink] New post   10 Sep 2012, 00:04
ShalabhsQuants wrote:
SOURH7WK wrote:
Min value of f(x,y) = 0, If we choose either x or y as 0 and the other as 25. The product x^2y^3 becomes 0.
Max value of f(x,y) = Infinity. Since x+y =25, and both are integers, we can choose y as a +ve number such as 10000xxxxxx000 and x as -ve no such as 999999xxxxx75. If we add we will get 25, but if we square x it will be a positive expression and the value of f(x,y) can go up to infinity.


As far as Minimum value of f(x,y) is concerned, your answer is correct. But Maximum value is not infinity. Remember x, & y both are non-negative integers. Minimum values of x, & y would be 0 only, whereas maximum as 25.


Ok, I missed that "non" part in the stem. Now looking at the choices (i,e ending with zeros) I tried few combination and got 10^2x15^3 = 33750. So is Option C is the maximum value?
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Re: The function f(x,y) is such that f(x,y) = x^2.y^3 & x+y =25. Where x, [#permalink] New post   10 Sep 2012, 01:54
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SOURH7WK wrote:
ShalabhsQuants wrote:
SOURH7WK wrote:
Min value of f(x,y) = 0, If we choose either x or y as 0 and the other as 25. The product x^2y^3 becomes 0.
Max value of f(x,y) = Infinity. Since x+y =25, and both are integers, we can choose y as a +ve number such as 10000xxxxxx000 and x as -ve no such as 999999xxxxx75. If we add we will get 25, but if we square x it will be a positive expression and the value of f(x,y) can go up to infinity.


As far as Minimum value of f(x,y) is concerned, your answer is correct. But Maximum value is not infinity. Remember x, & y both are non-negative integers. Minimum values of x, & y would be 0 only, whereas maximum as 25.


Ok, I missed that "non" part in the stem. Now looking at the choices (i,e ending with zeros) I tried few combination and got 10^2x15^3 = 33750. So is Option C is the maximum value?

You are right!
Let's understand the concept. This will eliminate the hit & trial approach for these kinds of questions.

If a function is such that f(a,b)=a^m.b^n, & a+b= constant, then f(a,b) would be maximum when a/m=b/n.

Coming to this question....

Given is x+y=25 =constant. for f(x,y)=x^2.y^3 to be max., x/2 should be equal to y/3 =>x/2=y/3 or x=2y/3.

By plugging in this value in x+y=25, we get 2y/3+y=25 => y=15, & x=10.

So Maximum of f(x,y)= 10^2.15^3 = 337500.
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Re: The function f(x,y) is such that f(x,y) = x^2.y^3 & x+y =25. Where x, [#permalink] New post   10 Sep 2012, 02:47
ShalabhsQuants wrote:

If a function is such that f(a,b)=a^m.b^n, & a+b= constant, then f(a,b) would be maximum when a/m=b/n.

Coming to this question....

Given is x+y=25 =constant. for f(x,y)=x^2.y^3 to be max., x/2 should be equal to y/3 =>x/2=y/3 or x=2y/3.

By plugging in this value in x+y=25, we get 2y/3+y=25 => y=15, & x=10.

So Maximum of f(x,y)= 10^2.15^3 = 337500.


How you have derived that formula. The formula seems very conditional with a+b constant & only for maximum value. Is there any partial derivative involved??
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Re: The function f(x,y) is such that f(x,y) = x^2.y^3 & x+y =25. Where x, [#permalink] New post   10 Sep 2012, 05:13
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SOURH7WK wrote:
ShalabhsQuants wrote:

If a function is such that f(a,b)=a^m.b^n, & a+b= constant, then f(a,b) would be maximum when a/m=b/n.

Coming to this question....

Given is x+y=25 =constant. for f(x,y)=x^2.y^3 to be max., x/2 should be equal to y/3 =>x/2=y/3 or x=2y/3.

By plugging in this value in x+y=25, we get 2y/3+y=25 => y=15, & x=10.

So Maximum of f(x,y)= 10^2.15^3 = 337500.


How you have derived that formula. The formula seems very conditional with a+b constant & only for maximum value. Is there any partial derivative involved??


Formally, yes, it is by partial derivatives, looking for extremum point...

A sort of justification without partial derivatives:
For any real numbers \(x\) and \(y\), \(\, (x + y)^2 \geq{4xy}\). Equality holds if and only if \(x=y\) (the given inequality is equivalent to \((x-y)^2\geq{0}\). In words: when the sum of two real numbers is constant, the maximum product of the two numbers is obtained when they are each equal to half of the sum.

In our case, the sum is constant, but in the product we have two different powers, 2 and 3. Intuitively, the maximum will be obtained for a weighted average between \(x\) and \(y\), \(y\) being closer to 25 as in the product it has a greater power, but still not "too far away" from the half of the sum.

But, since here we have integers and in addition, it is a GMAT multiple choice question, we can use some number properties.
The possible answers (after we eliminate infinity) are all multiples of 5, and since the sum\(x+y=25\) is a multiple of 5, if one of the numbers is multiple of 5, then the other one is also. And of course, \(y\) should be greater than \(x.\)
Therefore, we only have to check \(x=5, \, y=20\) and \(x=10, \, y=15.\)
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Re: The function f(x,y) is such that f(x,y) = x^2.y^3 & x+y =25. Where x, [#permalink]
10 Sep 2012, 05:13
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