Maximizing Area of a Racetrack: What are the Dimensions for Optimal Performance?

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SUMMARY

The discussion focuses on optimizing the dimensions of a 400 km racetrack composed of two straight sides and semicircles at each end to maximize the enclosed area. The key equations involved are the area of the semicircle, given by \( \frac{1}{2} \pi r^2 \), and the area of the rectangle, represented as \( lw \). The relationship between the width of the rectangle and the radius of the semicircles is established as \( r = \frac{w}{2} \). The total perimeter constraint is expressed as \( 2l + \frac{\pi}{4} w^2 = 400 \).

PREREQUISITES
  • Understanding of optimization problems in calculus
  • Familiarity with the area formulas for semicircles and rectangles
  • Knowledge of perimeter constraints in geometric figures
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study optimization techniques in calculus, particularly for geometric shapes
  • Learn how to derive relationships between dimensions in constrained optimization problems
  • Explore the application of the Lagrange multiplier method for optimization
  • Investigate real-world applications of racetrack design and performance metrics
USEFUL FOR

Students preparing for calculus exams, mathematicians interested in optimization, and engineers involved in racetrack design and performance analysis.

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Homework Statement


The Question is:
"A 400km racetrack is to be built with two straight sides and semicricles at the ends. Find the dimensions of the track that encloses the maximum area."

The two long sides of the rectangle are written with >/= to 100m (each)
The straight side of the 2 semi circles is written with >/= to 20m (each)
(>/= means greater than or equal to, just in case)

Homework Equations



Area of semicircle = 1/2 * pr2
Area of rectangle = lw

The Attempt at a Solution


Well i tried but its always the beginning setting up of optimization problems that is the killer, the rest is always easy. It is a study Q for an upcoming test.
 
Physics news on Phys.org
Find a way to write the length of the straight section in terms of the radius of the end circles. To do this, you need to use the fact that the total perimeter is 400km.
 
In the semicircle ends, r= w/2 where w is the "width" of the rectangle. Since there are two ends, you really have one circle. The area you want to maximize is \pi r^2= \pi/2 w^2. As danago said, the circumference of the figure is 400 m so 2l+ \pi/4 w^2= 400.
 

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