Optimization with Constrained Function


by d=vt+1/2at^2
Tags: calculus, function, optimization, restriction
d=vt+1/2at^2
d=vt+1/2at^2 is offline
#1
Jun4-09, 11:33 AM
P: 9
1. The problem statement, all variables and given/known data
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.



2. Relevant equations
1000 = LW
C = 2L + W + 5W


3. 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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Mark44
Mark44 is offline
#2
Jun4-09, 12:33 PM
Mentor
P: 20,933
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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