I'm trying to solve the kind of linear equation below such that the sum of the unknowns is maximised

ban1ka1u 2022-07-13 Answered
I'm trying to solve the kind of linear equation below such that the sum of the unknowns is maximised, but have been unable to find the solution.
10 y 1 + 12 y 2 + 15 y 3 = 50
It may just be I am searching using the wrong terminology, if so direction of where to look would be greatly appreciated too.EDIT: To add some more information, the initial problem involved four equations.
y 1 = K 1 / x 1
y 2 = K 2 / x 2
y 3 = K 3 / x 3
x 1 + x 2 + x 3 = X
Where K and X values are given constants, and x and y values are unknown. I am trying to find a solution that maximises the sum of the y values.

It is from this problem that I simplified it to
K 1 y 1 + K 2 y 2 + K 3 y 3 = X
With the first equation in the initial question just being arbitrary values.
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Answers (1)

Elliott Gilmore
Answered 2022-07-14 Author has 10 answers
Your goal appears to be to maximize the function f ( x ) = y 1 + y 2 + y 3 with the constraint that 10 y 1 + 12 y 2 + 15 y 3 = 50. This can be solved with Legrange multipliers. In case you haven't heard of this, the idea is that the for any critical point of a function f ( x 1 , x 2 , ) constrained by the equation g ( x ) = K where K is a constant, there exists a value of λ where
fx1=λgx1; fx2=λgx2; fx3=λgx3; ⋯
We can use this to solve your optimization problem by creating this system of equations:
1 = λ 10 y 1 2 ; 1 = λ 12 y 2 2 ; 1 = λ 15 y 3 2
which simplifies to λ = y 1 2 10 = y 2 2 12 = y 3 2 15 or 60 λ = 6 y 1 2 = 5 y 2 2 = 4 y 3 2 . You can substitute this is to the original constraint to get 2.5 y 1 2 = 50 and y 1 = 10 . Therefore, λ = 1. This means that the solution is ( 10 , 2 3 , 15 )
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