# math help needed please help?

A bridge over the river Kwai forms an arch that is modeled by the function below. (Note this is -0.0274x^2 the decimal poofed.)

h(x) = -00274x^2 + 1.34033x

a) How wide is the bridge? (Assuming that the arch length is equal to the width)

b) Could a sailboat with an 18m mast sail under it? Why or why not state the reason mathematically.

### 3 Answers

- husoskiLv 71 month ago
I'd start by writing h(x) = -ax^2 + bx and only plug numbers in at the end. This usually saves me both time and aggravation. Factor out -ax:

h(x) = -ax(x - b/a)

... and it's clear that h(x) = 0 when either x=0 and x=b/a, and those are the bases of bridge on the two shores. The difference between those two x values is b/a, and *that's* what I'd call the width of the bridge.

The graph of y = h(x) is a parabola, with x = b/(2a) -- vertical and halfway between the roots -- as its axis of symmetry. That make the maximum height:

h(b / (2a)) = -a (b/(2a))^2 + b (b/(2a)) = -b^2/(4a) + b^2/(2a) = b^2 / (4a)

Now you can plug in values for a and b. Notice that I've made a>0, so the minus sign is baked into the algebra.

If you really need arc length, that's a slightly messy integral, involving either a hyperbolic trig substitution or a creative integration by parts to get a solution in terms of natural logarithms instead of inverse hyperbolic sines.

- 冷眼旁觀Lv 61 month ago
a)

At the two end points, height = 0

When h(x) = 0:

-00274x² + 1.34033x = 0

x (-00274x + 1.34033) = 0

x = 0 or x = 1.34033/0.0274

x = 0 or x = 48.9 (to 3 sig. fig.)

Width of the bridge = 48.9 m

====

b)

h(x) = -0.0274x² + 1.34033x

h(x) = -0.0274 [x² - 48.9x]

h(x) = -0.0274 [x² - 48.9x + (48.9/2)²] + 0.0274 * (48.9/2)²

h(x) = -0.0274 (x - 24.45)² + 16.4

For all real values of x, (x - 24.45)² ≥ 0, and thus -0.0274 (x - 24.45)² ≤ 0

h(x) = -0.0274 (x - 24.45)² + 16.4 ≤ 16.4

Maximum height of the bridge = 16.4 m

The mast is higher than the height of the bridge.

Hence, the sailboat with an 18 m mast could NOT sail under the bridge.