If {a^2}{x^4} + {b^2}{y^4} = {c^6},then the m | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation ${a^2}{x^4} + {b^2}{y^4} = {c^6}$.Using the $AM$-$GM$ inequality,we know that for positive numbers $u$ and $v$,$\frac{u+v}{2} \

Vedclass · Accepted Answer

$\frac{{{c^3}}}{{\sqrt {2ab} }}$

Vedclass · Answer

(C) Given the equation ${a^2}{x^4} + {b^2}{y^4} = {c^6}$.Using the $AM$-$GM$ inequality,we know that for positive numbers $u$ and $v$,$\frac{u+v}{2} \ge \sqrt{uv}$.Let $u = {a^2}{x^4}$ and $v = {b^2}{y^4}$.Then $\frac{{{a^2}{x^4} + {b^2}{y^4}}}{2} \ge \sqrt{{a^2}{x^4} \cdot {b^2}{y^4}}$.Substituting the given equation: $\frac{{{c^6}}}{2} \ge \sqrt{{a^2}{b^2}{x^4}{y^4}}$.$\frac{{{c^6}}}{2} \ge ab{x^2}{y^2}$.${x^2}{y^2} \le \frac{{{c^6}}}{{2ab}}$.Taking the square root on both sides,we get $xy \le \sqrt{\frac{{{c^6}}}{{2ab}}}$.Therefore,the maximum value of $xy$ is $\frac{{{c^3}}}{{\sqrt {2ab} }}$.

If ${a^2}{x^4} + {b^2}{y^4} = {c^6}$,then the maximum value of $xy$ is

Similar Questions

If $a, b > 0$,then the minimum value of $y = \frac{b^2}{a-x} + \frac{a^2}{x}$ for $0 < x < a$ is

The range of $a \in R$ for which the function $f(x)=(4 a-3)\left(x+\log _{e} 5\right)+2(a-7) \cot \left(\frac{x}{2}\right) \sin ^{2}\left(\frac{x}{2}\right)$ where $x \neq 2 n \pi, n \in N$,has critical points,is

The abscissa of the points,where the tangent to the curve $y=x^3-3x^2-9x+5$ is parallel to the $X$-axis,are

If $m$ is the minimum value of $k$ for which the function $f(x) = x\sqrt{kx - x^2}$ is increasing in the interval $[0, 3]$ and $M$ is the maximum value of $f$ in $[0, 3]$ when $k = m$,then the ordered pair $(m, M)$ is equal to

The points on the curve $y = 12x - x^3$ at which the gradient is zero are

Vedclass Test Series

Exam Paper Generator

Online Exam Module