A tunnel is built in form of a parabola. The width at the base of tunnel is 7 m. On
the ceiling 3 m from each end of the base there are light fixtures. The height to light
fixtures is 4 m. A trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive
through the tunnel? Is this possible? (Give full algebraic solution.

Given:

The width at the base of parabolic tunnel is 7 m.

The ceiling 3 m from each end of the base there are light fixtures.

The height to light fixtures is 4 m.

To find:

Whether it is possible a trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel.

Solution:

The width at the base of tunnel is 7 m.

Let the graph of the parabola intersect the x-axis at x=0 and x=7. It means x and (x-7) are the factors of the height function.

The function of height is:

[tex]h(x)=ax(x-7)[/tex]             ...(i)

Where, a is a constant.

The ceiling 3 m from each end of the base there are light fixtures and the height to light fixtures is 4 m. It means the graph of height function passes through the point (3,4).

Putting x=3 and h(x)=4 in (i), we get

[tex]4=a(3)((3)-7)[/tex]

[tex]4=a(3)(-4)[/tex]

[tex]\dfrac{4}{(3)(-4)}=a[/tex]

[tex]-\dfrac{1}{3}=a[/tex]

Putting [tex]a=-\dfrac{1}{3}[/tex], we get

[tex]h(x)=-\dfrac{1}{3}x(x-7)[/tex]              ...(ii)

The center of the parabola is the midpoint of 0 and 7, i.e., 3.

The width of the truck is 4 m. If is passes through the center then the truck must m 2 m on the left side of the center and 2 m on the right side of the center.

2 m on the left side of the center is x=1.5.

A trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is possible if h(1.5) is greater than 2.8.

Putting x=1.5 in (ii), we get

[tex]h(1.5)=-\dfrac{1}{3}(1.5)(1.5-7)[/tex]

[tex]h(1.5)=-(0.5)(-5.5)[/tex]

[tex]h(1.5)=2.75[/tex]

It is clear that h(1.5)<2.8, therefore the trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is not possible.

Is algebra.
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