If a+b=n, then what is the maximum value of ab? (do not use calculus to solve)?

Question

also, what values of a and b give the maximum product?

Answer 1

Assume n is even, n = 2m, say
choices
a - - b ab
1 (n - 1) (n -1)
2 (n - 2) 2(n - 2)
3 (n - 3) 3 (n - 3)
...
...
m m m^2 = n^2/4 - - - max for n = 2m

similarly, if n is odd n = 2m+1, say

a = m, b = m+1, ab = m(m+1) is max

Answer 2

For ab to be maximized, a and b should both be the same sign, because if one were negative then ab would be negative. This question is equivalent to asking what shape rectangle has the greatest area, which is of course a square. So a and b should be equal to achieve the greatest product. Since a + b = n, a and b must both be equal to n/2. Therefore, the maximum value of ab is n²/4.
There are two answers for the values of a and b that give the maximum product.
a = b = n/2
and
a = b = -n/2

Answer 3

ab = (1/4) [ (a + b)^2 - (a - b)^2 ]

If a and b are positive, then ab is maximum when a - b on RHS above is zero.

So, maximum value of ab = n^2/4

Also a = b when ab is maximum gives
a = b = n/2

Answer 4

By the Arithmetic Mean - Geometric Mean Inequality (AM-GM), we get the following:

(a+b)/2 >= sqrt(ab).

Squaring both sides yields

((a+b)/2)^2 >= ab.

ab attains maximum value when there is equality so max(ab) = ((a+b)^2)/4 = (n^2)/4. It reaches this max value when a = b = n/2.

Answer 5

Let's see what happens if a and b are different from n/2

Then there's a number x such that a=(n/2 - x) and
b = (n/2 + x), thus

ab = (n/2 + x)(n/2 - x) = (n^2)/4 - x^2 < (n^2)/ 4 = (n/2)*(n/2)

If a=b=n/2 ab = (n/2)^2 = (n^2)/4

Thus, (n^2/4) is the maximum.

Answer 6

Ummm...seeing as there are no numbers and only variables involved, the anwser for everything would be infinity.

In terms of variables though, it would be an-a^2.
Work time.
a+b=n b=n-a ab=a(n-a) = an-a^2