We can solve this problem using the equation of an ellipse. Since the tunnel has a semi-elliptical shape, the equation of the ellipse will be:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Where:

• $a$ is the semi-major axis (half the width of the tunnel).
• $b$ is the semi-minor axis (the height of the tunnel).
• $x$ represents the horizontal distance from the center of the ellipse.
• $y$ represents the height of the ellipse at that point.

Given Information:

• The tunnel is 33 feet wide, so $a = \frac{33}{2} = 16.5$ feet.
• The tunnel is 27 feet high, so $b = 27$ feet.
• The vehicle is 6 feet wide, meaning its horizontal distance from the center to one side is $x = \frac{6}{2} = 3$ feet.

We want to find the maximum height $y$ of the vehicle at this distance from the center.

Step-by-Step Calculation:

1. Use the equation of the ellipse at $x = 3$:

$\frac{3^2}{16.5^2} + \frac{y^2}{27^2} = 1$

2. Solve for $y$:

$\frac{9}{272.25} + \frac{y^2}{729} = 1$

$0.033 + \frac{y^2}{729} = 1$

$\frac{y^2}{729} = 1 - 0.033 = 0.967$

$y^2 = 0.967 \times 729 = 704.343$

$y = \sqrt{704.343} \approx 26.54 \text{ feet}$

Thus, the maximum height of the vehicle should be approximately 26.54 feet to safely drive under the archway.

Would you like more details or have any questions?

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Here are five related questions for deeper exploration:

1. What would happen if the vehicle width increased to 8 feet?
2. How does the width of the road affect the safe passage of vehicles under the archway?
3. What is the maximum height of a vehicle that can pass through a circular tunnel with the same width and height?
4. How does the shape of the tunnel (e.g., elliptical vs. rectangular) affect the clearance height for vehicles?
5. If the archway were flattened, how would that impact the required vehicle height?

Tip: When solving geometric problems involving ellipses, always visualize the shape and apply the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ for precision.