Question

Two vertical poles of heights 60 ft and 80 ft stand on level ground, with their bases 100 ft apart. A cable that is stretched from the top of one pole to some point on the ground between the poles, and then to the top of the other pole. What is the minimum possible length of cable required

Answer 1

Answer:

Step-by-step explanation:

Given

height of two Poles are 60 and 80 ft

Distance between them is 100 ft

Let x be the distance of Pole of ht 80 ft from Point of stretch

thus length of rope is given by

$L=L_1+L_2$

$L_1=\sqrt{80^2+x^2}$

$L_2=\sqrt{60^2+(100-x)^2}$

$L=\sqrt{80^2+x^2}+\sqrt{60^2+(100-x)^2}$

differentiate w.r.t x we get

$\frac{\mathrm{d} L}{\mathrm{d} x}=\frac{2x}{2\sqrt{80^2+x^2}}-\frac{2\left ( 100-x\right )}{\sqrt{60^2+\left ( 100-2x\right )^2}}$

Put $\frac{\mathrm{d} L}{\mathrm{d} x}=0$ to get minimum value

$\frac{2x}{2\sqrt{80^2+x^2}}=\frac{2\left ( 100-x\right )}{\sqrt{60^2+\left ( 100-2x\right )^2}}$

squaring

$x^2\left ( (100-x)^2+60^2\right )=(100-x)^2(80^2+x^2)$

Rearranging

$28x^2-1800x+640000=0$

$x=\frac{400}{7}$

thus $L_1=98.312 ft$

$L_2=73.73 ft$

$L=172.04 ft$