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The least value of the function f(x) = abax+bx (where a > 0, b > 0, x > 0) is ______.

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Question

The least value of the function f(x) = `"a"x + "b"/x` (where a > 0, b > 0, x > 0) is ______.

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Solution

The least value of the function f(x) = `"a"x + "b"/x` (where a > 0, b > 0, x > 0) is `2sqrt("ab")`.

Explanation:

Here, f(x) = `"a"x + "b"/x`

⇒ f'(x) = `"a" - "b"/x^2`

For maximum and minimum value f'(x) = 0

∴ `"a" - "b"/x^2` = 0

⇒ `x^2 = "b"/"a"`

⇒ x = `+- sqrt("b"/"a")`

Now f"(x) = `(2"b")/x^3`

`"f''"(x)_(x = sqrt("b"/"a")) = (2"b")/(("b"/"a")^(3/2))`

= `2 ("a"^(3/2))/("b"^(1/2)) > 0`

Hence, minima

So the least value of the function at x = `sqrt("b"/"a")` is 

`"f"(sqrt("b"/"a")) = "a" * sqrt("b"/"a") + "b"/sqrt("b"/"a")`

= `sqrt("ab") + sqrt("ab")`

= `2sqrt("ab")`

Hence, least value is `2sqrt("ab")`.

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Chapter 6: Application Of Derivatives - Exercise [Page 142]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 6 Application Of Derivatives
Exercise | Q 64 | Page 142

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