Question

Find the minimum value of (ax + by), where xy = c².

Answer 1

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 \[2c\sqrt{ab}\].

Answer 2

Given that xy = c²

`y = c^2/x`   ...(i)

Now, suppose S = ax + by

⇒ `S = ax + b xx c^2/x`   ...[From (i)]

⇒ `"dS"/"dx" = a - (bc^2)/x^2`

For local points of maxima or minima

⇒ `"dS"/"dx" = 0`

⇒ `a - (bc^2)/x^2 = 0`

⇒ `x = - c sqrt(b/a)`

Also, ``

`(d^2S)/(dx^2)]_("at"  x  =  csqrt(b/a)) = (2bc^2)/(c^3(b/a)^(3//2)) > 0`

∴ S = ax + by is minimum at `x = csqrt(b/a)`

⇒ Minimum value of `S = a xx csqrt(b/a) + b xx c^2/(csqrt(b/a))`

= `csqrt(ab) + csqrt(ab)`

= `2csqrt(ab)`

∴ Minimum value of ax + by, where xy = c² is `2csqrt(ab)`.