How far from the taller tower will the wire touch the ground if the length of the wire is a minimum?Two towers 40m apart are 30m and 20m high. A wire at the top of each tower is guyed to the ground...

How far from the taller tower will the wire touch the ground if the length of the wire is a minimum?

Two towers 40m apart are 30m and 20m high. A wire at the top of each tower is guyed to the ground between them. 

Asked on by kiahreid

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Begin with a segment representing the 40m distance between the towers. At one endpoint erect a perpendicular segment representing the 30m tall tower, while on the other endpoint erect a perpendicular segment representing the 20m tall tower. Pick a point X on the 40m segment where the guy wires meet.

Then two right triangles have been formed -- one has legs of 30m and x meters; the other has legs 20m and (40-x)meters.

Since we are to minimize the total length of the guy wires, we find the lengths of the hypotenuse for each triangle and add them together.

If `l_1` is the length of the hypotenuse of the triangle with the 30m leg, we have: `l_1=sqrt(30^2+x^2)=sqrt(900+x^2)`

If `l_2` is the length of the hypotenuse of the other triangle then `l_2=sqrt(20^2+(40-x)^2)=(x^2-80x+2000)`

The total length of the guy wire is `D=sqrt(900+x^2)+sqrt(x^2-80x+2000)`

To minimize `D(x)` we take the first derivative and examine the critical points.

`D'(x)=1/2(900+x^2)^(-1/2)(2x)+1/2(x^2-80x+2000)^(-1/2)(2x-80)`

`=x/sqrt(900+x^2)+(x-40)/(sqrt(x^2-80x+2000))`  Setting `D'(x)=0` we get:

`(40-x)/sqrt(x^2-80x+2000)=x/sqrt(900+x^2)` Then:

`(40-x)sqrt(900+x^2)=xsqrt(x^2-80x+2000)` squaring both sides:

`(x^2-80x+1600)(900+x^2)=x^2(x^2-80x+2000)`

`1440000+1600x^2-72000x-80x^3+900x^2+x^4=x^4-80x^3+2000x`

`500x^2-72000x+1440000=0`

`x^2-144x+2880=0`

`(x-24)(x-120)=0` so x=24 or x=120. x=120 does not work in the context of the problem, so x=24 is the critical point we seek. It is a minimum (check using the first derivative rule, or recognize the graph is a parabola opening up)

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The distance from the taller tower is 24m

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** This problem is much easier using basic geometry. The angle where the guy wire meets the ground will be the same when the total length is minimum, so the triangles are similar.

`30/x=20/(40-x)==>50x=1200==>x=24`

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