Find two integers whose difference is 12 and whose product is a minimum.
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The difference of the two integers is 12. Let one of them be A, the other is 12 + A.
We have to minimize A*(12 + A)
The value of A for which P = A^2 + 12A is the minimum can be found by solving P' = 0 for A and ensuring that P''(A) is positive.
P' = 2A + 12
2A + 12 = 0
=> A = -6
P'' = 2
At A = -6, P'' is positive, this indicates the minimum value is at A = -6
The integers are A = -6 and A + 12 = 6
The required integers are -6 and 6
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6 and -6
Let a be the larger one of two integers, b the other one, such that a -b =12. let m = a -6 = b + 6, the average of (a + b)/2. The product
a b = (m + 6) (m -6) = m^2 -12 >= -12. m^12 - 12 = -12 only when m = 0, the minimal point of m^2 -12.
from m =0, we get a = 6, b = -6.
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