Answer:
The rope should be placed at 1.46 from the 16 ft pole to minimize the length.
Step-by-step explanation:
From the diagram, our goal is to minimize the length of Rope AC passing through B.
First, we determine the length of the rope AC.
AC=AB+BC
In the first triangle,
[tex]|AB|^2=16^2+x^2\\|AB|=\sqrt{16^2+x^2}[/tex]
Similarly, in the second triangle,
[tex]|BC|^2=24^2+(30-x)^2\\|BC|=\sqrt{x^2-60x+1476}[/tex]
Length of the Rope, AC
[tex]L=\sqrt{16^2+x^2}+\sqrt{x^2-60x+1476}[/tex]
First, to minimize L,we find its derivative.
[tex]L'=\dfrac{x\sqrt{x^2-60x+1476}+(x-30)\sqrt{16^2+x^2}}{(\sqrt{16^2+x^2})(\sqrt{x^2-60x+1476})}[/tex]
Setting the derivative to zero
[tex]x\sqrt{x^2-60x+1476}+(x-30)\sqrt{16^2+x^2}=0\\-x\sqrt{x^2-60x+1476}=(x-30)\sqrt{16^2+x^2}\\$Square both sides\\x^2(x^2-60x+1476)=(x-30)^2(16^2+x^2)\\x^4-60x^3+1476x^2=x^4-60x^3+1156x^2-15360x+230400\\1476x^2=1156x^2-15360x+230400\\1476x^2-1156x^2+15360x-23040=0\\320x^2+15360x-23040=0\\x=1.46,-49.46[/tex]
The rope should be placed at 1.46 from the 16 ft pole to minimize the length.
The least amount of the rope is the smallest length that can be gotten from the pole
The rope should be placed at 12 feet from the 16 ft pole to use the least amount of rope.
The heights are given as:
[tex]\mathbf{h_1 = 16}[/tex]
[tex]\mathbf{h_2 = 24}[/tex]
The distance is given as:
[tex]\mathbf{d = 30}[/tex]
See attachment for the illustrating diagram
Considering the two right-angled triangles on the diagram, we have the following equations, using Pythagoras theorem
[tex]\mathbf{L_1 = \sqrt{x^2 + 16^2}}[/tex]
[tex]\mathbf{L_2 = \sqrt{(30 - x)^2 + 24^2}}[/tex]
Expand
[tex]\mathbf{L_1 = \sqrt{x^2 + 256}}[/tex]
[tex]\mathbf{L_2 = \sqrt{900 - 60x +x^2 + 576}}[/tex]
[tex]\mathbf{L_2 = \sqrt{1476- 60x +x^2 }}[/tex]
The length (L) of the pole is:
[tex]\mathbf{L = L_1 + L_2}[/tex]
So, we have:
[tex]\mathbf{L = \sqrt{x^2 + 256} + \sqrt{1476 - 60x + x^2}}[/tex]
Differentiate
[tex]\mathbf{L' = \frac{x}{\sqrt{x^2 + 256}} + \frac{x - 30}{\sqrt{1476 - 60x + x^2}}}[/tex]
Set to 0
[tex]\mathbf{\frac{x}{\sqrt{x^2 + 256}} + \frac{x - 30}{\sqrt{1476 - 60x + x^2}} = 0}[/tex]
Take LCM
[tex]\mathbf{\frac{x\sqrt{1476 - 60x + x^2} +(x - 30)\sqrt{x^2 + 256}}{\sqrt{x^2 + 256} \times \sqrt{1476 - 60x + x^2}} = 0}[/tex]
Simplify
[tex]\mathbf{x\sqrt{1476 - 60x + x^2} +(x - 30)\sqrt{x^2 + 256} = 0}[/tex]
Rewrite as:
[tex]\mathbf{x\sqrt{1476 - 60x + x^2} =-(x - 30)\sqrt{x^2 + 256} }[/tex]
Square both sides
[tex]\mathbf{x^2(1476 - 60x + x^2) =(x^2 - 60x + 900)(x^2 + 256) }[/tex]
Expand
[tex]\mathbf{1476x^2 - 60x^3 + x^4 =x^4 - 60x^3 + 900x^2 + 256x^2 - 15360x + 230400}[/tex]
Simplify
[tex]\mathbf{1476x^2 - 60x^3 + x^4 =x^4 - 60x^3 + 1156x^2 - 15360x + 230400}[/tex]
Evaluate like terms
[tex]\mathbf{1476x^2 = 1156x^2 - 15360x + 230400}[/tex]
Rewrite as:
[tex]\mathbf{1476x^2 - 1156x^2 + 15360x - 230400 = 0}[/tex]
[tex]\mathbf{320x^2 + 15360x - 230400 = 0}[/tex]
Divide through by 320
[tex]\mathbf{x^2 + 48x - 720 = 0}[/tex]
Using a calculator, we have:
[tex]\mathbf{x = \{12,-60\}}[/tex]
The value of x cannot be negative.
So, we have:
[tex]\mathbf{x = 12}[/tex]
Hence, the rope should be placed at 12 feet from the 16 ft pole
Read more about minimizing lengths at:
https://brainly.com/question/15174196
