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max2010maxim [7]

3 years ago

14

A Russian blue kitten costs $150 more than Manx kitten 9 blue kittens and 8 Manx kittens costs. Total of 14,100 write down a alg

ebraic equation in the simplest form

1 answer:

Dmitriy789 [7]3 years ago

5 0

Answer:

The equations 9x+8y=14100 and x = y+150 are in the simplest form.

Step-by-step explanation:

Given that:

Russian blue kitten costs $150 more than Manx kitten.

Cost of 9 blue kittens and 8 Manx kittens = $14100

Let,

x be the price of one Russian blue kitten.

y be the price of one Manx kitten.

According to given statement;

9x+8y=14100

x = y+150

Hence,

The equations 9x+8y=14100 and x = y+150 are in the simplest form.

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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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lutik1710 [3]

<u>Anna:</u>

1st month = 18 + $\frac{3}{4}$ = 18 $\frac{3}{4}$ inches

2nd month = 18 $\frac{3}{4}$ + $\frac{3}{4}$ = 19 $\frac{1}{2}$

3rd month = 19 $\frac{1}{2}$ + $\frac{3}{4}$ = 20 $\frac{1}{4}$

4th month = 20 $\frac{1}{4}$ + $\frac{3}{4}$ = 21

5th month = 21 + $\frac{3}{4}$ = 21 $\frac{3}{4}$

<u>Grace:</u>

Because Anna is $\frac{9}{4}$ inches shorter than Grace, we can start the count from the third month:

1st month = 20 $\frac{1}{4}$

2nd month = 21

3rd month = 21 $\frac{3}{4}$

4th month = 21 $\frac{3}{4}$ + $\frac{3}{4}$ = 22 $\frac{1}{2}$

5th month = 22 $\frac{1}{2}$ + $\frac{3}{4}$ = 23

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Step-by-step explanation:

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