NIMCET 2011 Question Paper with solutions

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Question : 41 of 120

Marks: +1, -0

The minimum value of

p x+q y

, when

x y=r^{2}

is

$2 r √{p q}$
$2 p q √{r}$
$-2 r √{p q}$
$√{p q r}$

Solution:

Given,

x y=r^{2}

...(i)
Let

u=p x+q y,

[from Eq. (i)]

⇒ u=p x+q( {r^{2}}/{x})

⇒ {d u}/{d x}=p- {q r^{2}}/{x^{2}}

For maxima and minima, put

{d u}/{d x}=0

p- {q r^{2}}/{x^{2}}=0 ⇒ x=± r √{ {q}/{p}}

Now,

{d^{2} u}/{d x^{2}}= {2 q r^{2}}/{x^{3}}

At

x=r {√{q}}/{ρ}

{d^{2} u}/{d x^{2}}>0

, minima
Minimum value is

u=p × r √{ {q}/{p}}+q( {r^{2}}/{r √{ {q}/{p}}})

=p r √{ {q}/{p}}+q r √{ {p}/{q}}

=r √{p q}+r √{p q}=2 r √{p q}

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