Expert: Ahmed Salami - 11/16/2008

Question
The sum of the first and twice the second is 100 and the product is a maximum.

Answer
Hi Shanice,
The first and second terms of an AP are a and a+d where d is the common difference.
If the sum of the first and twice the second is 100, then
a + 2(a+d) = 100
a + 2a + 2d = 100
3a + 2d = 100
2d = 100 - 3a
d = (100 - 3a)/2
The product of the first and the second is
P = a(a+d)
 = a^2 + ad
 = a^2 + a[(100 - 3a)/2]
 = a^2 + a(50 - 3a/2)
 = a^2 + 50a - 3a^2/2
 = 50a - a^2/2
To find the maximum value of P, we differentiate with respect to a and equate to zero.
dP/da = 50 - a
at dP/da = 0, 50 - a = 0
a = 50
d = (100 - 3a)/2
 = (100 - 150)/2
 = -50/2
 = -25
Therefore, P is maximum when a = 50 and d = -25

Regards.