Answer: amount of wire needed is 5.84 feet.
Step-by-step explanation:
given data:
height of the first pole = 8 feet
height of the second pole = 10 feet
distance between both poles = 50 feet
SOLUTION:
L1 = [tex]\sqrt{x^{2} + 8^{2} } \\[/tex]
= [tex]\sqrt{x^{2} + 64}[/tex]
L2 = [tex]\sqrt{(50-x)^{2} +10^{2} }[/tex]
= [tex]\sqrt{(50-x)^{2}+100 }[/tex]
total length of wire
L = L1 + L2
L = [tex]\sqrt{x^{2} + 64}+\sqrt{(50-x)^{2}+100 }[/tex]
[tex]\frac{dL}{dx}=\frac{2x}{2\sqrt x^{2} +64} + \frac{2(50-x(-1)}{2\sqrt{(50-x)^{2} +100} } =0[/tex]
[tex]\frac{x}{\sqrt x^{2} +64} = \frac{50-x}{\sqrt{(50-x)+100} }[/tex]
[tex]x\sqrt{(50-x} )^{2}+100 = (50-x)\sqrt{x^{2} +64}[/tex]
[tex]x^{2} ((50-x)+100)=(15-x)^{2} (x^{2} +64)[/tex]
[tex]x^{2} (50-x)^{2} +100x^{2} =x^{2} (50-x)^{2} +64(50-x)^{2} \\\\100x^{2} = 64(50-x)^{2} \\100x^{2} =64((50-x)(50-x))\\100x^{2} = 64(2500-100x+x^{2} )\\\\100^{2} =160000 -6400x+64x^{2} \\\\36x^{2} +6400x-160000=0[/tex]
[tex]36x^{2} + 64x-1600=0[/tex]
[tex]18x^{2} +32x+800=0[/tex]
using quadratic equation
[tex]a=18\\b=32\\c=800\\[/tex]
refer to the attached image
[tex]x=\frac{-8}{9} +\frac{4}{9} \sqrt{229} \\or \\x=x=\frac{-8}{9} +\frac{-4}{9} \sqrt{229}[/tex]
[tex]x = 5.84\\or\\x= -7.61[/tex]
