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Andru [333]
3 years ago
14

Please help ❗️❗️❗️ I will 100% mark brainly

Mathematics
2 answers:
nevsk [136]3 years ago
6 0
A1=8
a2=12
a3=18
a4=27
a5=40.5
a6=60.75
Lina20 [59]3 years ago
3 0

Answer:

Step-by-step explanation:

a1=8

a2=12

a3=18

a4=27

a5=40.5

a6=60.75

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4 is the x1 value, -8 is the y2 value, 8 is the x2 value, and 5 is the y2 value.

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In the function y = 10.00x + 35, y represents the cost of cabin lodging per occupant and x represents the number of occupants. H
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Y = 10x + 35

if y is cost per occupant and x is occupants, then we can read the equation as such:

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Lagrange multipliers have a definite meaning in load balancing for electric network problems. Consider the generators that can o
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Answer:

The load balance (x_1,x_2,x_3)=(545.5,272.7,181.8) Mw minimizes the total cost

Step-by-step explanation:

<u>Optimizing With Lagrange Multipliers</u>

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

\displaystyle C_i=3x_i+\frac{i}{40}x_i^2

It means the cost for each generator is expanded as

\displaystyle C_1=3x_1+\frac{1}{40}x_1^2

\displaystyle C_2=3x_2+\frac{2}{40}x_2^2

\displaystyle C_3=3x_3+\frac{3}{40}x_3^2

The total cost of production is

\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

Simplifying and rearranging, we have the objective function to minimize:

\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)

The restriction can be modeled as a function g(x)=0:

g: x_1+x_2+x_3=1000

Or

g(x_1,x_2,x_3)= x_1+x_2+x_3-1000

We now construct the auxiliary function

f(x_1,x_2,x_3)=C(x_1,x_2,x_3)-\lambda g(x_1,x_2,x_3)

\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)

We find all the partial derivatives of f and equate them to 0

\displaystyle f_{x1}=3+\frac{2}{40}x_1-\lambda=0

\displaystyle f_{x2}=3+\frac{4}{40}x_2-\lambda=0

\displaystyle f_{x3}=3+\frac{6}{40}x_3-\lambda=0

f_\lambda=x_1+x_2+x_3-1000=0

Solving for \lambda in the three first equations, we have

\displaystyle \lambda=3+\frac{2}{40}x_1

\displaystyle \lambda=3+\frac{4}{40}x_2

\displaystyle \lambda=3+\frac{6}{40}x_3

Equating them, we find:

x_1=3x_3

\displaystyle x_2=\frac{3}{2}x_3

Replacing into the restriction (or the fourth derivative)

x_1+x_2+x_3-1000=0

\displaystyle 3x_3+\frac{3}{2}x_3+x_3-1000=0

\displaystyle \frac{11}{2}x_3=1000

x_3=181.8\ MW

And also

x_1=545.5\ MW

x_2=272.7\ MW

The load balance (x_1,x_2,x_3)=(545.5,272.7,181.8) Mw minimizes the total cost

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