Using maxima and minima, the minimum possible value of their sum is [tex]2\sqrt{k}[/tex].
Let the two numbers be a and b,
So, it is given that the product of two positive numbers, a, and b, is a fixed number k.
So, a*b = k
b = k/a
Now, S = a + b
S = a + k/a
dS/da = 0
dS/da = [tex]1 - ka^{-2}[/tex]
So,
[tex]1 - ka^{-2} = 0\\\\ka^{-2} =1 \\\\\frac{k}{a^{2} } = 1\\\\k = a^{2} \\\\k = \sqrt{a}[/tex]
[tex]a = \sqrt{k}[/tex]
Now, Calculating the second derivative,
[tex]\frac{d^{2}S }{da^{2} } = 2\frac{k}{a^{3} }[/tex]
Now, putting [tex]a = \sqrt{k}[/tex] in the second derivative, we get
[tex]\frac{d^{2}S }{da^{2} } = 2\frac{k}{\sqrt{k} ^{3} } = \frac{2}{\sqrt{k} }[/tex] which is greater than 0.
So, for [tex]a = \sqrt{k}[/tex] and [tex]b = \sqrt{k}[/tex]
we will have their sum to be minimum.
So, the minimum value of a + b is
[tex]a + b = \sqrt{k} +\sqrt{k} =2\sqrt{k}[/tex]
Hence, the minimum possible value of their sum is [tex]2\sqrt{k}[/tex].
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