Find the area described by $\displaystyle x+y+z=L$ and $\displaystyle 0<x,y,z<\frac{L}{2}$ Is this the SAME area as $\displaystyle x+y+z = \frac{L}{2}$ and $\displaystyle x,y,z>0$? Thanks~!
Follow Math Help Forum on Facebook and Google+
The first equation is off. x+y+z will be less than 3L/2 but not necessarily less than L/2
View Tag Cloud