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miss Akunina [59]

3 years ago

10

I will give brainiest to whoever answers correctly !!

1 answer:

Kisachek [45]3 years ago

4 0

I think the answer is $166,742.24

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

Ivahew [28]

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

5 0

3 years ago

What 2 numbers equal -80

mina [271]

-10 and 8 multiply them to equal 80 but if there both negative it will be a positive 80 because if you multiply 2 negatives its postive hope this helps

4 0

3 years ago

Read 2 more answers

6x7x6x9 this one is for you

Papessa [141]

Answer:HOPE THIS HELPS! :D

2268

Step-by-step explanation:

6x7=42

42x6=252

252x9=2268

so, 6x7x6x9=2268

8 0

3 years ago

in a fundraising committee of 90 people the ratio of men to women is 7:2. find the number of women required to join the existing

LuckyWell [14K]

Hello,

Step-by-step explanation:

90 = 9 x 10 = (7 + 2) x 10 = 70 + 20

70 men and 20 women

$\frac{70}{5} = \frac{20 + X}{4}$

56 = 20 + X ⇔ X = 36

Answer : + 36 women

7 0

3 years ago

What is the multiplicative inverse of c/d? Explain how a number is related to its multiplicative inverse.

Lana71 [14]

Answer:

$\frac{d}{c}$

Step-by-step explanation:

The multiplicative inverse of a number n is $\frac{1}{n}$

The product of the number and its multiplicative inverse = 1

n × $\frac{1}{n}$ = 1

Thus the multiplicative inverse of $\frac{c}{d}$ is

$\frac{1}{\frac{c}{d} }$ = $\frac{d}{c}$

6 0

3 years ago