Answer:
The load balance [tex](x_1,x_2,x_3)=(545.5,272.7,181.8)[/tex] Mw minimizes the total cost
Step-by-step explanation:
Optimizing With Lagrange Multipliers
When a multivariable function f is to be maximized or minimized, the Lagrange multipliers method is a pretty common and easy tool to apply when the restrictions are in the form of equalities.
Consider three generators that can output xi megawatts, with i ranging from 1 to 3. The set of unknown variables is x1, x2, x3.
The cost of each generator is given by the formula
[tex]\displaystyle C_i=3x_i+\frac{i}{40}x_i^2[/tex]
It means the cost for each generator is expanded as
[tex]\displaystyle C_1=3x_1+\frac{1}{40}x_1^2[/tex]
[tex]\displaystyle C_2=3x_2+\frac{2}{40}x_2^2[/tex]
[tex]\displaystyle C_3=3x_3+\frac{3}{40}x_3^2[/tex]
The total cost of production is
[tex]\displaystyle C(x_1,x_2,x_3)=3x_1+\frac{1}{40}x_1^2+3x_2+\frac{2}{40}x_2^2+3x_3+\frac{3}{40}x_3^2[/tex]
Simplifying and rearranging, we have the objective function to minimize:
[tex]\displaystyle C(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)[/tex]
The restriction can be modeled as a function g(x)=0:
[tex]g: x_1+x_2+x_3=1000[/tex]
Or
[tex]g(x_1,x_2,x_3)= x_1+x_2+x_3-1000[/tex]
We now construct the auxiliary function
[tex]f(x_1,x_2,x_3)=C(x_1,x_2,x_3)-\lambda g(x_1,x_2,x_3)[/tex]
[tex]\displaystyle f(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)-\lambda (x_1+x_2+x_3-1000)[/tex]
We find all the partial derivatives of f and equate them to 0
[tex]\displaystyle f_{x1}=3+\frac{2}{40}x_1-\lambda=0[/tex]
[tex]\displaystyle f_{x2}=3+\frac{4}{40}x_2-\lambda=0[/tex]
[tex]\displaystyle f_{x3}=3+\frac{6}{40}x_3-\lambda=0[/tex]
[tex]f_\lambda=x_1+x_2+x_3-1000=0[/tex]
Solving for \lambda in the three first equations, we have
[tex]\displaystyle \lambda=3+\frac{2}{40}x_1[/tex]
[tex]\displaystyle \lambda=3+\frac{4}{40}x_2[/tex]
[tex]\displaystyle \lambda=3+\frac{6}{40}x_3[/tex]
Equating them, we find:
[tex]x_1=3x_3[/tex]
[tex]\displaystyle x_2=\frac{3}{2}x_3[/tex]
Replacing into the restriction (or the fourth derivative)
[tex]x_1+x_2+x_3-1000=0[/tex]
[tex]\displaystyle 3x_3+\frac{3}{2}x_3+x_3-1000=0[/tex]
[tex]\displaystyle \frac{11}{2}x_3=1000[/tex]
[tex]x_3=181.8\ MW[/tex]
And also
[tex]x_1=545.5\ MW[/tex]
[tex]x_2=272.7\ MW[/tex]
The load balance [tex](x_1,x_2,x_3)=(545.5,272.7,181.8)[/tex] Mw minimizes the total cost
