Chapter 3.6, Problem 45E

To determine

To calculate: The distances x for which the arch is above the road.

Answer to Problem 45E

The arch is above the road from 55.213 meters from pylon to 447.156 meters from the same pylon.

Explanation of Solution

Given information:

The model of the arch: $y=-0.00211x^2+1.06x$

Distance from the left pylons = x meters

Height of the arch above the water: $y=52\text{meters}$

Formula used:

Perimeter of a rectangle:

Perimeter of a rectangle is $2\left(\text{length}\times \text{width}\right)$

Area of the rectangle:

Area of a rectangle is $\text{length}\times \text{width}$

Calculation:

The given model of the arch is $y=-0.00211x^2+1.06x$ .

The height of the arch above the water is $y=52$ .

Substitute 52 for y in the given equation.

  $\begin{array}{c}52=-0.00211x^2+1.06x\\ 0.00211x^2-1.06x+52=0\end{array}$

Solve the above equation using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Substitute 0.00211 for a , −1.06 for b and 52 for c in the above formula.

  $\begin{array}{c}x=\frac{-\left(-1.06\right)\pm \sqrt{{\left(-1.06\right)}^2-4\left(0.00211\right)\left(52\right)}}{2\left(0.00211\right)}\\ x=\frac{1.06\pm \sqrt{1.1236-0.43888}}{0.00422}\\ x\approx \frac{1.06\pm 0.827}{0.00422}\end{array}$

Solve for x .

  $\begin{array}{c}x\approx \frac{1.06+0.827}{0.00422}\\ \approx \frac{1.887}{0.00422}\\ \approx 447.156\end{array}$

And, $\begin{array}{c}x\approx \frac{1.06-0.827}{0.00422}\\ \approx \frac{0.233}{0.00422}\\ \approx 55.213\end{array}$

Hence, the arch is above the road from 55.213 meters from pylon to 447.156 meters from the same pylon.