Q:

The product of two positive numbers is 1024. what is the minimum value of their sum?

Accepted Solution

A:
Let the two positive numbers are x , y and there sum is s
So,
            x y = 1024 ⇒⇒⇒⇒ (1)
And        S = x + y ⇒⇒⇒⇒ (2)
by substituting from (1) at (2) with the value of y = 1024/x

∴ s = x + [tex] \frac{1024}{x} [/tex]

Differentiating both sides with respect to x to find the minimum value of the sum and equating to zero

∴  [tex] \frac{ds}{dx} = 1 - \frac{1024}{ x^{2} } = 0[/tex]

solve the last equation for x
∴ [tex]1 - \frac{1024}{ x^{2} } = 0[/tex]
    [tex] 1 = \frac{1024}{ x^{2} } [/tex]  ⇒⇒⇒ *x²
∴ x²  = 1024
∴ [tex] x = \pm \sqrt{1024} = \pm 32[/tex]
taking the positive number as stated in the problem

x = 32

So, The numbers are 32 , 32
And theire mimimum sum = 32 + 32 = 64