Question

Find the greatest value of $x^3{y}^4$ if $2x+3y=7$ and $x\ge 0,y\ge 0$.

Answer 1

$2x+3y=7\therefore x=\frac{7-3y}{2}{x}^{3}x{y}^4=(\frac{7-3y}{2}{)}^{3}x{y}^4=f(y)\therefore f^{{}^{\prime }}(y)=\frac{d}{dy}[\frac{y^4}{8}(7-3y{)}^3]f^{{}^{\prime }}\left(y\right)=\frac{4y^3}{8}(7-3y{)}^3+\frac{y^4}{8}\times 3(7-3y{)}^2(-3)$

For extrreme values $f^{{}^{\prime }}\left(y\right)=0\Rightarrow \frac{4y^3}{8}(7-3y{)}^3+y\frac{4}{8}\times 3(7-3y{)}^2(-3)=0=0,\frac{28}{21}{f}^{{}^{\prime }}\left(\frac{28}{21}\right)<0$

$\therefore$ maxima at$\frac{28}{21}y=\frac{28}{21}\Rightarrow x=\frac{3}{2}\therefore x^3{y}^4=\left(\frac{3}{2}{)}^3\right(\frac{28}{21}{)}^4=\frac{32}{3}$

is greater value