Optimization with Constrained Function

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SUMMARY

The discussion focuses on optimizing the cost of fencing for a 1000m² garden with specific constraints. The garden has three sides made of wooden fence and one side made of vinyl, which costs five times more than wood. The relationship between length (L) and width (W) is constrained such that L cannot exceed 30% greater than W. The cost function is defined as C = 2L + W + 5W, and the solution involves using calculus to minimize costs while adhering to the given constraints.

PREREQUISITES
  • Understanding of algebraic manipulation and equations
  • Knowledge of calculus, specifically optimization techniques
  • Familiarity with cost functions and constraints in mathematical modeling
  • Basic geometry related to area calculations
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  • Study optimization techniques in calculus, focusing on constrained optimization
  • Learn about cost function formulation in mathematical modeling
  • Explore algebraic methods for solving equations with constraints
  • Investigate practical applications of optimization in real-world scenarios
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Students in mathematics, particularly those studying calculus and optimization, as well as professionals involved in cost analysis and resource allocation in project management.

d=vt+1/2at^2
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Homework Statement


1000m^2 garden. 3 sides made of wooden fence. 1 side made of vinyl(costs 5x as much as wood).

Length cannot be more than 30% greater than the width.

Find the dimensions for the minimum cost of the fence.



Homework Equations


1000 = LW
C = 2L + W + 5W


The Attempt at a Solution


Attempted ignoring the restriction. Answer does not meet restriction. Solved algebraically for the only rectangle where L = 1.3W and L = 1000/W. It is a calculus question and it is therefore suspected that this is not the answer. The minimum cost is not necessarily when the vinyl side is minimal.
 
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I would start it this way: Solve the equation LW = 1000 for one variable, say W. Then write your cost function as a function of W alone. Use calculus techniques to find the minimum cost over the interval that includes all possible values of W, given the constraint that the length can't exceed 130% of the width.
 

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