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bogdanovich [222]
3 years ago
11

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex

treme values of the function subject to the given constraint. f(x1, x2, ..., xn) = x1 + x2 + ... + xn; x12 + x22 + ... + xn2 = 4
Mathematics
1 answer:
olga nikolaevna [1]3 years ago
6 0

f(x_1,\ldots,x_n)=x_1+\cdots+x_n=\displaystyle\sum_{i=1}^nx_i

{x_1}^2+\cdots+{x_n}^2=\displaystyle\sum_{i=1}^n{x_i}^2=4

The Lagrangian is

L(x_1,\ldots,x_n,\lambda)=\displaystyle\sum_{i=1}^nx_i+\lambda\left(\sum_{i=1}^n{x_i}^2-4\right)

with partial derivatives (all set equal to 0)

L_{x_i}=1+2\lambda x_i=0\implies x_i=-\dfrac1{2\lambda}

for 1\le i\le n, and

L_\lambda=\displaystyle\sum_{i=1}^n{x_i}^2-4=0

Substituting each x_i into the second sum gives

\displaystyle\sum_{i=1}^n\left(-\frac1{2\lambda}\right)^2=4\implies\dfrac n{4\lambda^2}=4\implies\lambda=\pm\frac{\sqrt n}4

Then we get two critical points,

x_i=-\dfrac1{2\frac{\sqrt n}4}=-\dfrac2{\sqrt n}

or

x_i=-\dfrac1{2\left(-\frac{\sqrt n}4\right)}=\dfrac2{\sqrt n}

At these points we get a value of f(x_1,\cdots,x_n)=\pm2\sqrt n, i.e. a maximum value of 2\sqrt n and a minimum value of -2\sqrt n.

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The triangles below are similar. Triangle A B C. Side A C is 10 and side A B is 5. Angle C is 30 degrees. Triangle D E F. Side E
Gnoma [55]

Answer:

\triangle ABC {\displaystyle \sim } \triangle DE\ F

\triangle CBA {\displaystyle \sim } \triangle FED

\triangle BAC {\displaystyle \sim } \triangle EDF

Step-by-step explanation:

Given

\triangle ABC {\displaystyle \sim } \triangle DE\ F

AC = 10

AB = 5

\angle C = 30^\circ

ED = 7.5

DF = 25

\angle F =30

\angle E =90

Required

Which of the options is/are true:

\triangle CBA {\displaystyle \sim } \triangle FED            \triangle CBA {\displaystyle \sim } \triangle FDE              \triangle BAC {\displaystyle \sim } \triangle EFD

\triangle BAC {\displaystyle \sim } \triangle EDF            \triangle ABC {\displaystyle \sim } \triangle DE\ F             \triangle ABC {\displaystyle \sim } \triangle DFE

The given triangles, implies that:

A {\displaystyle \sim } \ D

B {\displaystyle \sim } \ E

C {\displaystyle \sim } \ F

By taking each sides of both triangle, one after the other; the possible similar triangles are:

\triangle ABC {\displaystyle \sim } \triangle DE\ F

\triangle CBA {\displaystyle \sim } \triangle FED

\triangle BAC {\displaystyle \sim } \triangle EDF

5 0
3 years ago
PLEASE HELP step by step
Katen [24]

Answer:

The length of the line segment is 10 units.

Step-by-step explanation:

Let us call the given two points: (x₁, y₁) = (2, -3) and (x₂, y₂) = (8, 5).

The distance between two points, d = $ \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} } $

Here, $ (x_1, y_1) = (2, -3) $ and

$ (x_2, y_2) = (8, 5) $

Therefore, distance, d = $ \sqrt{(8 - 2)^2 + (5 - (-3))^2 $

$ = \sqrt{6^2 + 8^2} $

$ = \sqrt{36 + 64} $

$ = \sqrt{100} $

= 10 units.

Hence, the answer.

3 0
3 years ago
Clare subscribes to an online music streaming service
xeze [42]

Answer:

She should go with service 2 because it is, cheaper. Service 1 cost $107.52, but service 2 cost only $105.

Step-by-step explanation:

Cost now+increase$=new cost

(Part=percent*whole)

Increase=percent increase*cost now

i=12%*$96

i=0.12*96=$11.52

$96+11.52=107.52

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2 years ago
Hey all!
Rainbow [258]

Answer:

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Step-by-step explanation:

<u>Use the law of cosines:</u>

  • cos C = (25² + 27² - 28²)/(2*25*27)
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8 0
3 years ago
How to solve a quadratic sequence
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6 0
3 years ago
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