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konstantin123 [22]
3 years ago
9

Whats the first 5 multiples of 13 and all factors

Mathematics
2 answers:
zavuch27 [327]3 years ago
4 0

Answer:

first 5 multiples are 13,26,39,52,65,78 and factors are 1 and 13

Step-by-step explanation:

solniwko [45]3 years ago
3 0

The first 5 multiples of 13 are 13, 26, 39, 52, and 65. The factors of 13 are 1 and 13

<u>Solution:</u>

Given that we need to find the first 5 multiples of 13 and factors of 13

<em><u>Finding first 5 multiples of 13:</u></em>

13 \times 1 = 13\\13 \times 2 = 26\\13 \times 3 = 39\\13 \times 4 = 52\\13 \times 5 = 65

Thus the first 5 multiples of 13 are 13, 26, 39, 52 and 65

<em><u>Finding factors of 13:</u></em>

The number 13 is a prime number. Because 13 is only divided by 1 and by itself(13)

Hence the factors of 13 are 1 and 13

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<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.

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\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

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

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

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

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x_1=3x_3

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

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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

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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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