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A=1/2πw^2+2lw

?????????????

User Linehrr
by
7.6k points

1 Answer

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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

x=(-b+/-√(b^2-4ac))/(2a)
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
User Samer Makary
by
7.7k points
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