1 Answer

Samer Makary · Answer 1 · 2018-03-03T10:16:09+0000

alrighty, solve for w

minus A from both sides
multiply both sides by 2
2A=πw^2+4lw
minus 2A from both sides

$0=\pi w^2+4lw-2A$
factor
can't
use quadratic formula
for
0=ax^2+bx+c

so for

$0=\pi w^2+4lw-2A$
a=pi
b=4l
c=-4A

$w=\frac{-(4l)+/-√((4l)^2-4(\pi)(-4A)){2(\pi)}$

$w=\frac{-4l+/-√(16l^2+16A\pi){2\pi}$

$w=\frac{-4l+/-√(16(l^2+A\pi)){2\pi}$

$w=\frac{-4l+/-4√(l^2+A\pi){2\pi}$

$w=\frac{-2l+/-2√(l^2+A\pi){\pi}$

so therfor,

$w=\frac{-2l+2√(l^2+A\pi){\pi}$ or
$w=\frac{-2l-2√(l^2+A\pi){\pi}$
evaluate each one and whihever one is negative, ignore that solution

0 Comments

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

0 Comments

Please log in or register to add a comment.

Related questions

Other Questions