I need to find dimensions of rectangular box open at the top , of maximum capacity whose Surface area is 432 sq.units
Now I have to do this using Lagrange multipliers so I write as
$F(x,y,z,\lambda) = xyz + \lambda(2yz+2zx+xy)$
$F_x = yz + \lambda(2z+y) = 0 $
$F_y = xz + \lambda(2z+x)=0$
$F_z=xy+\lambda(2y+2x)=0$
from 1 and 2nd equation I get $x=y$. but how do I proceeed ?
Thanks
Volume $V$ of box = $xyz$ and constraint : $$2xy + 2xz + xy = 432$$
Using Lagrange multipliers one can write that $$ F(x,y,z, \lambda) = xyz + \lambda ( 2xz + 2yz + xy- 432)$$ For stationary points, partial derivatives should be zero. $$ \begin{align} \Rightarrow F_{\lambda} &= 2xy + 2yz + xy - 432 = 0 \\ \Rightarrow F_x &= yz + \lambda(2z+y) = 0 \\ \Rightarrow F_y &= xz + \lambda(2z+x)=0 \\ \Rightarrow F_z &=xy+\lambda(2y+2x)=0 \end{align} $$
Which on solving gives two stationary points $(0,0,0)$ and $(12,12,6)$, it should be obvious that former cannot be the maximum, hence for $(12,12,6)$ volume would be maximum.