Find the sides of the right triangle so that their sum is minimized

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SUMMARY

The discussion centers on finding the sides of a right triangle that minimize the sum of the legs \(a + b\) given a fixed hypotenuse \(c = 5\). The mathematical approach involves using the Pythagorean theorem \(a^2 + b^2 = c^2\) and calculus to derive that \(a = \frac{c}{\sqrt{2}} \approx 3.54\) and \(b = \frac{c}{\sqrt{2}} \approx 3.54\) yield a maximum sum of approximately 7.07. However, the minimum sum occurs when either \(a\) or \(b\) approaches zero, resulting in a sum of 5, which is not feasible for a triangle. The conclusion is that there is no valid minimum for \(a + b\) under the constraints of triangle geometry.

PREREQUISITES
  • Understanding of the Pythagorean theorem
  • Basic calculus, specifically differentiation
  • Knowledge of constraints in optimization problems
  • Familiarity with triangle properties and definitions
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  • Study the implications of constraints in optimization problems
  • Explore the concept of limits and their application in calculus
  • Learn about integer solutions in Pythagorean triples
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Mathematicians, students studying geometry and calculus, educators teaching optimization problems, and anyone interested in the properties of triangles and their applications in real-world scenarios.

  • #31
Hill said:
What would be enough is assuming that North-South directions and East-West directions are mutually orthogonal everywhere in the area in question.
This is better.
Hill said:
If one starts at any point on that map and goes South, he follows a ray which goes from the North Pole and passes through the starting point.
If he goes East or West, he follows a circle which has its center at The North Pole and which passes through the starting point.
Thank you one more time for your help.
 
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  • #32
MatinSAR said:
It's not enough for the question you mentioned, But I guess it's enough to solve the main question which is related to high school. Do you agree? He doesn't walk that much(only 5 km) so the question doesn't want us to consider that he is walking on a sphere and ...
It does not really matter that @Hill disagrees.

The given problem, as you eventually stated in post #9, is quite clear. Your recent replies to @FactChecker reinforce that the given problem refers to 2D grid on a planar surface.
 
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  • #33
SammyS said:
It does not really matter that @Hill disagrees.

The given problem, as you eventually stated in post #9, is quite clear. Your recent replies to @FactChecker reinforce that the given problem refers to 2D grid on a planar surface.
Thank you @SammyS for your help.

Thank you to everyone who helped for his time.
 

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