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

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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Elliott Gilmore
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 $\left(\sqrt{10},2\sqrt{3},\sqrt{15}\right)$