Question

A bridge on a river is modeled by the equation h = -0.2d^2 + 2.25d, where h is the height and d is the horizontal distance. For cleaning and maintenance purposes a worker wants to tie a taut rope on two ends of the bridge so that he can slide on the rope. The rope is at an angle defined by the equation -d + 6h = 21.77. If the rope is attached to the bridge at points A and B, such that point B is at a higher level than point A, at what distance from the ground level is point A?

Answer 1

Hello,
A=(2.21082279...,3.996803798)

`\left \{ {{h=-0.2d^2+2.25d} \atop {h=d/6+21.77/6}} \right. \\\\ \Rightarrow\ 1.2d^2-12.5d+21.77=0\\\\ \Rightarrow\ d=2.21082279..\ or\ d=8.20584..\\\\`

Answer 2

Since you are interested in values of h, it is convenient to eliminate d from the equations. The second equation tells you
d = 6h - 21.77

Substituting that into the first equation gives
h = (6h -21.77)(-0.2(6h -21.77) +2.25) = (6h -21.77)(-1.2h +6.604)
7.2h² -64.748h +143.76908 = 0 . . . . rearranging to standard form

The quadratic formula tells you the solution to
ax²+bx+c=0
is ...
x = (-b±√(b²-4ac))/(2a)

Using this formula on our quadratic, we have
h = (64.748 ±√((-64.748)² -4(7.2)(143.76908)))/(2·7.2)
h = (64.748 ±√51.754)/14.4
h ≈ 3.997 or 4.996

Point A is 3.997 from ground level.