Problem Solving

[Math Revolution GMAT math practice question]

If x and y are positive integers and xy=36, what is the smallest possible value of x+y?

A. 6
B. 8
C. 10
D. 12
E. 18

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When we know the product of two numbers, the sum of those numbers is a minimum when the two numbers are equal.

Since xy = 36, the minimum value of x + y occurs when x = y = 6. It is 6 + 6 = 12.

Therefore, D is the answer.
Answer: D

Max@Math Revolution wrote: When we know the product of two numbers, the sum of those numbers is a minimum when the two numbers are equal.

This is an immediate consequence of the famous Arithmetic-Geometric Mean inequality , therefore out-of-scope of the GMAT.

Important detail: the condition about integers (given in the question stem) is unnecessary, but the condition on positive (in fact nonnegative) numbers is essential.

For the interested readers:

[Arithmetic-Geometric Mean inequality] wrote: Given any positive integer \[N \geqslant 2\] and \[{x_1}\,,\,\,{x_2}\,,\, \ldots ,\,\,{x_N}\] nonnegative real numbers , we have:
\[\sqrt[{N\,}]{{{x_1}\, \cdot \,\,{x_2} \cdot \, \ldots \cdot \,\,{x_N}}}\,\,\, \leqslant \,\,\,\frac{{{x_1}\, + \,\,{x_2}\, + \, \ldots + \,{x_N}}}{N}\]
with equality if, and only if, all real numbers considered are equal.

In our case, \[N = 2\] and we have \[\sqrt {36} = \sqrt {xy} \leqslant \,\,\frac{{x + y}}{2}\] and (x+y) will be minimum exactly when (x+y)/2 is minimum...

This occurs (by the famous theorem above) only when x and y are equal, i.e., when \[6 = \sqrt {36} = \sqrt {xy} = \sqrt {{x^2}} \,\,\,\mathop = \limits^{x\, \geqslant \,0} \,x\,\,\,\,\left( { = y} \right)\]

Therefore \[? = 6 + 6\] as mentioned.

I hope you find all this interesting.

Regards,
fskilnik.