Tunnel Graphing 380525

1. **State the problem:** Find the width of the tunnel from bottom left to bottom right using the quadratic model $$y = -0.08x^2 + 2.88x$$ where $$y$$ is the height and $$x$$ is the horizontal distance.\n\n2. **Formula and rules:** The width corresponds to the distance between the two points where the tunnel height $$y=0$$ (the x-intercepts). To find these, solve $$-0.08x^2 + 2.88x = 0$$.\n\n3. **Solve for x-intercepts:**\n$$-0.08x^2 + 2.88x = 0$$\nFactor out $$x$$:\n$$x(-0.08x + 2.88) = 0$$\nSet each factor to zero:\n$$x=0$$ or $$-0.08x + 2.88=0$$\nSolve second equation:\n$$-0.08x = -2.88$$\n$$x = \frac{-2.88}{-0.08} = \frac{2.88}{0.08}$$\nCalculate:\n$$x = 36$$\n\n4. **Answer:** The tunnel width is from $$x=0$$ to $$x=36$$ feet, so the width is $$36$$ feet.\n\n---\n\n5. **State the problem:** Find how close the tallest vehicle (max height 13.5 ft) can drive from the bottom left or right.\n\n6. **Formula:** Set $$y=13.5$$ and solve for $$x$$ in $$-0.08x^2 + 2.88x = 13.5$$.\n\n7. **Solve quadratic:**\n$$-0.08x^2 + 2.88x - 13.5 = 0$$\nMultiply entire equation by $$-1$$ to simplify:\n$$0.08x^2 - 2.88x + 13.5 = 0$$\nUse quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a=0.08$$, $$b=-2.88$$, $$c=13.5$$:\nCalculate discriminant:\n$$b^2 - 4ac = (-2.88)^2 - 4(0.08)(13.5) = 8.2944 - 4.32 = 3.9744$$\nSquare root:\n$$\sqrt{3.9744} \approx 1.9936$$\nCalculate roots:\n$$x = \frac{2.88 \pm 1.9936}{2 \times 0.08} = \frac{2.88 \pm 1.9936}{0.16}$$\nFirst root:\n$$x_1 = \frac{2.88 - 1.9936}{0.16} = \frac{0.8864}{0.16} = 5.54$$\nSecond root:\n$$x_2 = \frac{2.88 + 1.9936}{0.16} = \frac{4.8736}{0.16} = 30.46$$\n\n8. **Answer:** The tallest vehicle can drive as close as about $$6$$ feet from the bottom left or $$30$$ feet from the bottom right (rounded to nearest foot).\n\n---\n\n9. **State the problem:** Check if tunnel width $$36$$ ft is enough for two or three lanes, each lane is $$12$$ ft wide.\n\n10. **Calculate lane widths:**\nTwo lanes: $$2 \times 12 = 24$$ ft\nThree lanes: $$3 \times 12 = 36$$ ft\n\n11. **Answer:** Tunnel width $$36$$ ft is enough for two lanes (needs 24 ft) and exactly enough for three lanes (needs 36 ft).\n\n---\n\n12. **State the problem:** Calculate walkway space if 2 ft walls are built on each side of the road and two lanes are in the middle.\n\n13. **Calculate space used by lanes and walls:**\nTwo lanes: $$24$$ ft\nTwo walls: $$2 \times 2 = 4$$ ft\nTotal road + walls: $$24 + 4 = 28$$ ft\n\n14. **Calculate remaining space for walkways:**\nTunnel width $$36$$ ft minus road + walls $$28$$ ft leaves $$8$$ ft\nDivide by 2 for each side:\n$$\frac{8}{2} = 4$$ ft per walkway\n\n15. **Answer:** Each walkway has $$4$$ feet of space.\n\n---\n\n16. **State the problem:** Check if walkway ceiling height is enough for an average pedestrian.\n\n17. **Calculate height at walkway edges:**\nWalkways start $$2$$ ft walls + $$24$$ ft lanes + $$4$$ ft walkway = $$30$$ ft from bottom left on one side, and $$6$$ ft from bottom left on the other side (since total width is 36 ft).\nWe check height at $$x=4$$ ft and $$x=32$$ ft (walkway centers).\nCalculate $$y$$ at $$x=4$$:\n$$y = -0.08(4)^2 + 2.88(4) = -0.08(16) + 11.52 = -1.28 + 11.52 = 10.24$$ ft\nCalculate $$y$$ at $$x=32$$:\n$$y = -0.08(32)^2 + 2.88(32) = -0.08(1024) + 92.16 = -81.92 + 92.16 = 10.24$$ ft\n\n18. **Answer:** The ceiling height at walkways is about $$10.24$$ ft, which is tall enough for an average pedestrian (usually about 6 ft tall).\n\n---\n\n19. **State the problem:** Should DOT accept or reject the tunnel design?\n\n20. **Answer:** The tunnel is wide enough for two lanes and walkways with walls, and tall enough for vehicles and pedestrians. Therefore, the DOT should accept the proposed design.