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3 years ago

1 point

1 answer:

6 0

<h3>The volume of a container with a radius of 4 centimeters and a height of 4 centimeters is 67.2 cubic centimeter</h3>

<em><u>Solution:</u></em>

<em><u>The volume of a container varies jointly with the square of its radius, r, and its height, h</u></em>

Therefore,

$v \propto r^2 h\\\\v = k \times r^2h -------- eqn\ 1$

<em><u>The container has a height of 10 centimeters, and radius of 6 centimeters, and a volume of 377 cubic centimeters</u></em>

Substitute v = 377 and h = 10 and r = 6 in eqn 1

$377 = k \times 6^2 \times 10\\\\377 = k \times 360\\\\k = 1.047 \approx 1.05$

<em><u>What is the volume of a container with a radius of 4 centimeters and a height of 4 centimeters?</u></em>

Substitute k = 1.05 and r = 4 and h = 4 in eqn 1

$v = 1.05 \times 4^2 \times 4\\\\v = 1.05 \times 16 \times 4\\\\v = 67.2$

Thus volume of a container with a radius of 4 centimeters and a height of 4 centimeters is 67.2 cubic centimeter

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3 years ago

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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

I don’t know the answer please help !

Shalnov [3]

Answer:

1. 24

2. 24

3. -24

4. -24

I don`t know why some of the answers are 10, I think those are just trick ones. The answers are only in 24s

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Answer:

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<h2>HOPE IT HELPED U ...</h2>

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Anettt [7]

Answer:

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