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insens350 [35]
3 years ago
14

An arched drainage culvert can be modelled by the function:

Mathematics
1 answer:
Zanzabum3 years ago
5 0

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.



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