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

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?

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 \(x=10, \, y=15.\)

This is interesting concept. On further research, turns out this has been derived from weighted AM-GM concepts.

it is possible to solve the problem by basic maximum minimum problem using differentiation.

this problem can also be solved using basic calculus techniques (i.e., optimization with derivatives), even though the final answers need to be integers. However, on the GMAT derivatives are not tested.