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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