Question

please explain .... Ans batado jldi ​

Answer 1

Answer:

let

$y = {27}^{ \sin \alpha } \times {81}^{ \cos\alpha } \\ taking \: log \: both \: the \: sides \\ ln(y) = \sin( \alpha ) ln(27) \\ + \cos( \alpha ) ln(81) \\ now \: differentiating \:wrt \: x \\ \frac{d ln(y) }{dx} = \cos( \alpha ) ln(27) \\ \: \: \: \: \: \: \: \: \: \: \: \: - \sin( \alpha ) ln(81) \\ 0 \: \: \: \: \: \: \: \: = ( \cos( \alpha ) ln( {3}^{3} ) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - \sin( \alpha )ln( {3}^{4} ) )$

$\sin( \alpha ) ln( {3}^{4} ) = \cos( \alpha ) ln( {3}^{3} ) \\ \tan( \alpha ) = \frac{ ln( {3}^{3} ) }{ ln( {3}^{4} ) } = \frac{3 ln(3) }{4 ln(3) } = \frac{3}{4}$

$\tan( \alpha ) = \frac{p}{b} \\using \: pythagoras \: theorem \\ hypo. = \sqrt{ {p}^{2} + {b}^{2} } \\ hypo. = \sqrt{9 + 16} = 5$

$\cos( \alpha ) = \frac{b}{h} = + - \frac{4}{5}$

$\sin( \alpha ) = + - \frac{3}{5}$

put small value of sin theta and cos theta in your equation

we get

${3}^{3 \sin( \alpha ) } . {3}^{4 \cos( \alpha ) } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {3}(^{3 \times \frac{ - 3}{5} + 4 \times \frac{ - 4}{5} } )\\ \: \: \: \: \: \: \: \: \: \: \: \: \: = {3}^{ \frac{ - 9}{5} - \frac{16}{5} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = {3}^{ \frac{ - 25}{5} } = {3}^{ - 5}$

the minimum value of eq

${3}^{ - 5} = \frac{1}{243}$

the maximum value is

${3}^{3 sin( \alpha ) } . {3}^{4 \cos( \alpha ) } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {3}(^{3 \times \frac{ 3}{5} + 4 \times \frac{ 4}{5} } )\\ \: \: \: \: \: \: \: \: \: \: \: \: \: = {3}^{ \frac{ 9}{5} + \frac{16}{5} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = {3}^{ \frac{ 25}{5} } = {3}^{ 5} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = 243$

hope this will help u.......