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प्रश्न
Find the minimum value of (ax + by), where xy = c2.
उत्तर
Let z = ax + by .....(1)
Given:
xy = c2 or \[y = \frac{c^2}{x}\]
Putting
\[y = \frac{c^2}{x}\] in (1), we get
z = ax + \[\frac{b c^2}{x}\]
Differentiating both sides w.r.t. x, we get
\[\frac{dz}{dx} = a - \frac{b c^2}{x^2}\]
For maxima or minima,
\[\frac{dz}{dx} = 0\]
⇒ \[a - \frac{b c^2}{x^2} = 0\]
⇒ \[x^2 = \frac{b c^2}{a}\]
⇒ \[x = \pm c\sqrt{\frac{b}{a}}\]
Now,
\[\frac{d^2 z}{d x^2} = \frac{2b c^2}{x^3}\]
At \[x = c\sqrt{\frac{b}{a}}\] , \[\frac{d^2 z}{d x^2} = \frac{2b c^2}{\left( c\sqrt{\frac{b}{a}} \right)^3} > 0\]
\[\therefore x = c\sqrt{\frac{b}{a}}\] is the point of minima.
At \[x = - c\sqrt{\frac{b}{a}}\], \[\frac{d^2 z}{d x^2} = \frac{2b c^2}{\left( - c\sqrt{\frac{b}{a}} \right)^3} < 0\]
\[\therefore x = - c\sqrt{\frac{b}{a}}\] is the point of maxima.
So,
When \[x = c\sqrt{\frac{b}{a}}\], \[y = \frac{c^2}{x} = \frac{c^2}{c\sqrt{\frac{b}{a}}} = c\sqrt{\frac{a}{b}}\]
\[\therefore z_{\text { minimum}} = ac\sqrt{\frac{b}{a}} + bc\sqrt{\frac{a}{b}} = \frac{abc + abc}{\sqrt{ab}} = \frac{2abc}{\sqrt{ab}} = 2c\sqrt{ab}\]
Thus, the minimum value of (ax + by), where xy = c2 is \[2c\sqrt{ab}\].
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