If xy = c^2 , then minimum value of ax + by is - Vidyadip

MCQ

If $xy = {c^2},$ then minimum value of $ax + by$ is

A
$c\sqrt {ab} $
✓
$2c\sqrt {ab} $
C
$ - c\sqrt {ab} $
D
$ - 2c\sqrt {ab} $

✓

Answer

Correct option: B.

$2c\sqrt {ab} $

b
(b) $xy = {c^2}$ ==> $y = \frac{{{c^2}}}{x}$ ==> $f(x) = ax + by = ax + \frac{{b{c^2}}}{x}$

Differentiate with respect to $x$ $f'(x) = a - \frac{{b{c^2}}}{{{x^2}}}$

Put $f'(x) = 0$ ==> $a{x^2} - b{c^2} = 0$

==> ${x^2} = \frac{{b{c^2}}}{a}$ ==> $x = \pm \,\,c\sqrt {b/a} $

At $x = + \,c\sqrt {b/a,} \,\,ax + by$ will be minimum.

The minimum value $f\,\,\left( {c\sqrt {\frac{a}{b}} } \right) = a.c\sqrt {\frac{a}{b}} + \frac{{b{c^2}}}{c}.\sqrt {\frac{b}{a}} $

$= 2c\sqrt {ab} $.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If $\cos \theta + \sec \theta = \frac{5}{2}$, then the general value of $\theta $ is

The largest interval lying in $\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$ for which the function, $f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {\frac{x}{2} - 1} \right) + \log \left( {\cos x} \right)$ is defined is

The number of elements in the set $\left\{ n \in N : 10 \leq n \leq 100\right.$ and $3^{ n }-3$ is a multiple of $7\}$ is $........$.

Let $\alpha, \beta(\alpha>\beta)$ be the roots of the quadratic equation $x ^{2}- x -4=0$. If $P _{ a }=\alpha^{ n }-\beta^{ n }, n \in N$, then $\frac{ P _{15} P _{16}- P _{14} P _{16}- P _{15}^{2}+ P _{14} P _{15}}{ P _{13} P _{14}}$ is equal to$......$

If $A = \left[ {\begin{array}{*{20}{c}}2&2\\{ - 3}&2\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right],$ then ${({B^{ - 1}}{A^{ - 1}})^{ - 1}}$=

What is the equation of the locus of a point which moves such that $4$ times its distance from the $x$-axis is the square of its distance from the origin

Let $f : [-1,3] \to R$ be defined as

$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\left| x \right| + \left[ x \right],}&{ - 1 \leq x < 1} \\
{x + \left| x \right|,}&{1 \leq x < 2} \\
{x + \left| x \right|,}&{2 \leq x \leq 3}
\end{array}} \right.$

where $[t]$ denotes the greatest integer less than or equal to $t$. Then, $f$ is discontinuous at:

Domain of the function $\sqrt {2 - x} - \frac{1}{{\sqrt {9 - {x^2}} }}$ is

The number of $4$ digit even numbers that can be formed using $0, 1, 2, 3, 4, 5, 6$ without repetition is

If $g(x) = 2f (2x^3 - 3x^2) + f(6x^2 - 4x^3 - 3)$, $\forall x \in R$ and $f"(x) > 0, \forall x \in R$ , then $g'(x) > 0$ for $x$ belonging to