alrighty, solve for w
minus A from both sides
multiply both sides by 2
2A=πw^2+4lw
minus 2A from both sides
[tex]0=\pi w^2+4lw-2A[/tex]
factor
can't
use quadratic formula
for [tex]0=ax^2+bx+c[/tex]
[tex]x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}[/tex]
so for
[tex]0=\pi w^2+4lw-2A[/tex]
a=pi
b=4l
c=-4A
[tex]w=\frac{-(4l)+/-\sqrt{(4l)^2-4(\pi)(-4A)}{2(\pi)}[/tex]
[tex]w=\frac{-4l+/-\sqrt{16l^2+16A\pi}{2\pi}[/tex]
[tex]w=\frac{-4l+/-\sqrt{16(l^2+A\pi)}{2\pi}[/tex]
[tex]w=\frac{-4l+/-4\sqrt{l^2+A\pi}{2\pi}[/tex]
[tex]w=\frac{-2l+/-2\sqrt{l^2+A\pi}{\pi}[/tex]
so therfor,
[tex]w=\frac{-2l+2\sqrt{l^2+A\pi}{\pi}[/tex] or [tex]w=\frac{-2l-2\sqrt{l^2+A\pi}{\pi}[/tex]
evaluate each one and whihever one is negative, ignore that solution
