# 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.
$\frac{10}{{y}_{1}}+\frac{12}{{y}_{2}}+\frac{15}{{y}_{3}}=50$
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
$\frac{{K}_{1}}{{y}_{1}}+\frac{{K}_{2}}{{y}_{2}}+\frac{{K}_{3}}{{y}_{3}}=X$
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 $f\left(x\right)={y}_{1}+{y}_{2}+{y}_{3}$ with the constraint that $\frac{10}{{y}_{1}}+\frac{12}{{y}_{2}}+\frac{15}{{y}_{3}}=50$. 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 $f\left({x}_{1},{x}_{2},\cdots \right)$ constrained by the equation $g\left(x\right)=K$ where $K$ is a constant, there exists a value of $\lambda$ where
fx1=λgx1; fx2=λgx2; fx3=λgx3; ⋯
We can use this to solve your optimization problem by creating this system of equations:
$1=-\lambda \frac{10}{{y}_{1}^{2}}$; $1=-\lambda \frac{12}{{y}_{2}^{2}}$; $1=-\lambda \frac{15}{{y}_{3}^{2}}$
which simplifies to $-\lambda =\frac{{y}_{1}^{2}}{10}=\frac{{y}_{2}^{2}}{12}=\frac{{y}_{3}^{2}}{15}$ or $-60\lambda =6{y}_{1}^{2}=5{y}_{2}^{2}=4{y}_{3}^{2}$. You can substitute this is to the original constraint to get $2.5{y}_{1}^{2}=50$ and ${y}_{1}=\sqrt{10}$. Therefore, $\lambda =-1$. This means that the solution is $\left(\sqrt{10},2\sqrt{3},\sqrt{15}\right)$
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