Question

An arched drainage culvert can be modelled by the function:

Answer 1

Substitute 136.5 for y.

$136.5=\frac{x(191-x)}{60}$

$8190=x(191-x)$

$8190=191x-x^{2}$

$x^{2} -191x+8190=0$

Compare this equation with $ax^{2} +bx+c=0$

a = 1, b = - 191, c = 8190

$x=\frac{-b+\sqrt{b^{2}-4ac } }{2}$ or

$x=\frac{-b-\sqrt{b^{2}-4ac } }{2}$

$x=\frac{191+\sqrt{191^{2}-4(1)(8190) } }{2}$ or

$x=\frac{191-\sqrt{191^{2}-4(1)(8190) } }{2}$

$x=\frac{191+\sqrt{36481-32760 } }{2}$ or

$x=\frac{191-\sqrt{36481-32760 } }{2}$

$x=\frac{191+\sqrt{3721 } }{2}$ or

$x=\frac{191-\sqrt{3721 } }{2}$

$x=\frac{191+61}{2}$ or

$x=\frac{191-61}{2}$

$x=\frac{252}{2}$ or

$x=\frac{130}{2}$

x = 126 or x = 65

Hence, the points on the curve are (65, 136.5) and (126, 136.5).

The width of the air space is the distance between these points.

Width = $\sqrt{(x_{2}- x_{1})^{2} +( y_{2}- y_{1}) ^{2}$

= $\sqrt{(126- 65)^{2} +(136.5-136.5) ^{2}$

= 126 - 65

= 61

Hence, width of the air space is 61m.