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Korvikt [17]
3 years ago
7

Pls help for 27b!!!!!!!

Mathematics
1 answer:
rodikova [14]3 years ago
6 0

Step-by-step explanation:

\beta

is a root of the quadratic

4 {x}^{2}  - 10x + 5

This quadratic isn't factorable so use the quadratic formula

- b +  -  \sqrt{ \frac{b {}^{2}  - 4ac }{2a} }

The plus and minus sign after b means we going to have two roots.

The roots are

\frac{5 +  \sqrt{5} }{4}

and

\frac{5 -  \sqrt{5} }{4}

We subsitue those values for Beta.

27a.

\frac{5}{2}

27b.

25

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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
Allison has 1/2 cup of yogurt for making fruit parfait
ivolga24 [154]
Allison can make four parfaits because you need 1/8 cup of yogurt and Allison has 1/2 cup of yogurt. So the half of the cup would give her 4 * 1/8 to equal 1/2.
4 0
3 years ago
X/-4 = -1.1<br> I’m bad at one step equations so thank you if you help &lt;3
icang [17]

x/-4 = -1.1

=> x = -1.1 × -4

=> x = 4.4

8 0
3 years ago
Find the width of the rectangular prism when the surface area is 208 square centimeters? The height is 8cm and the length is 6cm
Blizzard [7]
SA = 2(wl + hl + hw)
SA = 208
h = 8
l = 6

208 = 2(6w + (8*6) + 8w)
208 = 2(14w + 48)
208 = 28w + 96
208 - 96 = 28w
112 = 28w
112/28 = w
4 = w <==== the width is 4 cm
6 0
3 years ago
Read 2 more answers
Simplify this 6r+3(2r+1)
SpyIntel [72]

Answer: 12r + 3

Step-by-step explanation:

6r+3(2r+1)

Apply BODMAS, do bracket first

6r + (6r + 3)

Do addition next

= 12r + 3

6 0
3 years ago
Read 2 more answers
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