a tunnel has a parabolic cross-section. it is 24m wide at the base, At a point 6m above the base, the width is 18m. what is the maximum height of the tunnel?
+129839 We know these points ( 12, 0) , (9,6) and the parabola will open downwards
And the verex will be at the max height, h......let the vertex = (0, h)
So we have
y = −ax^2 + h using the points we have
0 = −a(12)^2 + h → 0 = −144a + h (1)
6 = −a(9)^2 + h → 6 = −81a + h (2)
Subtract (1) from (2) and we have
6 = 63a solve for a
6/63 = a = 2/21
Find h using ( 1)
0 = −144(2/21) + h
144(2/21) = h
288/21 = h
h = 96/7 m ≈ 13.71 m
Here's a graph: https://www.desmos.com/calculator/a06alawmri
