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pantera1 [17]
2 years ago
8

Suppose a person is standing on the top of a building and that she has an instrument that allows her to

Mathematics
2 answers:
lions [1.4K]2 years ago
6 0

Answer:

Here is you annswer can you please give brainlist

Step-by-step explanation:

DerKrebs [107]2 years ago
4 0

Answer:

that is right

Step-by-step explanation:

yes I can do it

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Let production be given by P = bLαK1−α where b and α are positive and α < 1. If the cost of a unit of labor is m and the cost
Nana76 [90]

Answer:

The proof is completed below

Step-by-step explanation:

1) Definition of info given

We have the function that we want to maximize given by (1)

P(L,K)=bL^{\alpha}K^{1-\alpha}   (1)

And the constraint is given by mL+nK=p

2) Methodology to solve the problem

On this case in order to maximize the function on equation (1) we need to calculate the partial derivates respect to L and K, since we have two variables.

Then we can use the method of Lagrange multipliers and solve a system of equations. Since that is the appropiate method when we want to maximize a function with more than 1 variable.

The final step will be obtain the values K and L that maximizes the function

3) Calculate the partial derivates

Computing the derivates respect to L and K produce this:

\frac{dP}{dL}=b\alphaL^{\alpha-1}K^{1-\alpha}

\frac{dP}{dK}=b(1-\alpha)L^{\alpha}K^{-\alpha}

4) Apply the method of lagrange multipliers

Using this method we have this system of equations:

\frac{dP}{dL}=\lambda m

\frac{dP}{dK}=\lambda n

mL+nK=p

And replacing what we got for the partial derivates we got:

b\alphaL^{\alpha-1}K^{1-\alpha}=\lambda m   (2)

b(1-\alpha)L^{\alpha}K^{-\alpha}=\lambda n   (3)

mL+nK=p   (4)

Now we can cancel the Lagrange multiplier \lambda with equations (2) and (3), dividing these equations:

\frac{\lambda m}{\lambda n}=\frac{b\alphaL^{\alpha-1}K^{1-\alpha}}{b(1-\alpha)L^{\alpha}K^{-\alpha}}   (4)

And simplyfing equation (4) we got:

\frac{m}{n}=\frac{\alpha K}{(1-\alpha)L}   (5)

4) Solve for L and K

We can cross multiply equation (5) and we got

\alpha Kn=m(1-\alpha)L

And we can set up this last equation equal to 0

m(1-\alpha)L-\alpha Kn=0   (6)

Now we can set up the following system of equations:

mL+nK=p   (a)

m(1-\alpha)L-\alpha Kn=0   (b)

We can mutltiply the equation (a) by \alpha on both sides and add the result to equation (b) and we got:

Lm=\alpha p

And we can solve for L on this case:

L=\frac{\alpha p}{m}

And now in order to obtain K we can replace the result obtained for L into equations (a) or (b), replacing into equation (a)

m(\frac{\alpha P}{m})+nK=p

\alpha P +nK=P

nK=P(1-\alpha)

K=\frac{P(1-\alpha)}{n}

With this we have completed the proof.

5 0
3 years ago
One number is 3 times a first number. A third number is 100 more than the first number. If the sum of the three numbers is 500,
Likurg_2 [28]
<span>Let the first number be x.
Then the second number is 3x.
Then the third number is (x+100).
Sum of the numbers=500

x+3x+(x+100)=500</span>
<span>4x+(x+100)-500=0
</span><span>5x=400</span>
<span>x=80
</span><span>The first number is 80.
The second number is 240.
The third number is 180.

We can check by adding all three of those numbers together: 80+240+180 = 500
</span>
Hope this helps! If you have any questions just ask!
8 0
3 years ago
Which expression is equal to (7 times 2)times 2
Lady bird [3.3K]
The answer is 28. 7x2=14 and then 14x2=28. Remember PEMDAS
7 0
3 years ago
Ray Cupple bought a basic car costing $26,500.00, with options costing $725.00. There is a 6% sales tax in his state and a combi
Alenkinab [10]
Given:
cost:                        26,500
option:                  <u>        725</u>
total                         27,225
tax  (27,225 * 6%)<u>     1,633.50</u>
total                         28,858.50
license & reg. fee <u>          50.00</u>
total cost               28,908.50  Choice D.

5 0
3 years ago
Read 2 more answers
Lynn is trying to determine how far away Student B is from the balloon. He decides to use the
kaheart [24]

Answer:

See Explanation

Step-by-step explanation:

<em>The question is incomplete as the image that illustrates the scenario is not given.</em>

<em>However, I can deduce that the question is about a right-angled triangle.</em>

<em>So, I will give a general explanation on how to find each of the side of the triangle, given a side and an angle.</em>

<em />

<u>For triangle A (solve for b)</u>

Using cosine formula.

\cos \theta = \frac{Adjacent}{Hypotenuse}

\cos 60= \frac{5}{b}

Make b the subject

b= \frac{5}{\cos 60}

<u>For triangle B (solve for b)</u>

Using cosine formula.

\sin \theta = \frac{Opposite}{Hypotenuse}

\sin 60= \frac{b}{5}

Make b the subject

b = 5\sin 60

<u>For triangle C (solve for b)</u>

Using cosine formula.

\tan \theta = \frac{Opposite}{Adjacent}

\tan 60= \frac{b}{5}

Make b the subject

b = 5\tan 60

3 0
3 years ago
Read 2 more answers
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