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Suppose a, b, and n are positive integers and a + b = n. For what values of a and b maximize ab?

 

The only way I know of maximizing ab is by drawing a table of values and comparing numbers. It seems though that if n = 2k, then the maximum is obtained when a = b = k. If n = 2k + 1, the maximum is obtain when a = k and b = k + 1. Is there an intuitive way of showing/deriving this though. I can't seem to think of anything.

Posted
  e(ho0n3 said:
Is there an intuitive way of showing/deriving this though. I can't seem to think of anything.
Yes, I believe there is, using Calculus.

 

We wish to maximize x \cdot y subject to the constraint x+y=n. Therefore, y=n-x, so our problem is equivalent to maximizing f(x)=x(n-x)=nx-x^2.

 

Taking the first derivative, (and noting that f(x) is a downward pointing parabola, and thus, has only one local extrema, a global maximum), we get that f'(x)=n-2x, therefore the local maxima is at x=n/2.

 

Thus, if x=2k (k an integer), the maxima is at x=k, and substituting x into the constraint equation implies that y=k as well. However, if n=2k+1, then the maxima is achieved at x = (2k+1)/2=k+1/2, which is not an integer. Therefore, the closest integers are x=k and x=k+1, which are both a distance of 1/2 away from x, and thus, both optimal integer solutions. Again, applying the constraint equation shows that these x-values correspond to y=k+1 and y=k respectively.

 

If you have any questions about my explanation, I will gladly expound any requested points.

Posted

I've found that, especially with things like GCSE maths problems. It's like trying to break a nut with a lorry as opposed to your average nutcracker ;)

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