Multivariable problems

goku900

New member
Joined
Mar 22, 2013
Messages
27
Ok the first problem is:

The output Q of an economic system subject to two inputs, such as labor L and capital K, s often modeled by the Cobb-Douglas production function Q(L;K) = cLaKb, where a; b and c are positive real numbers. When a+b = 1, the case is called constant returns to scale. Suppose
a = 1
3 , b = 2
3 and c = 40.

A) If L is held constant at L = 10, write the function that gives the dependence of Q on K.
B) If K is held constant at K = 15, write the function that gives the dependence of Q on L

\(\displaystyle Q(L,\, K)\, =\, cLaKb\)

\(\displaystyle \log Q(L,\,K)\, =\, \log c\, +\, \log L\, +\, \log a\, +\, \log K\, +\, \log b\)

\(\displaystyle \Rightarrow\, \dfrac{1}{Q} \dfrac{\partial Q}{\partial L}\, =\, \dfrac{1}{L}\). . . . .\(\displaystyle \dfrac{1}{Q} \dfrac{\partial Q}{\partial K}\, =\, \dfrac{1}{K}\)

. . . . .\(\displaystyle \dfrac{\partial Q}{\partial L}\, =\, \dfrac{Q}{L}\). . . . . ..\(\displaystyle \dfrac{\partial Q}{\partial K}\, =\, \dfrac{Q}{K}\)

. . . . ..\(\displaystyle Q_L\, =\, \dfrac{Q}{L}\). . . . . . .\(\displaystyle Q_K\, =\, \dfrac{Q}{K}\)

. . . . . . . .\(\displaystyle =\, \dfrac{cLaKb}{L}\). . . . . . .\(\displaystyle =\, \dfrac{cLaKb}{K}\)

. . . . . . . .\(\displaystyle =\, caKb\). . . . . . .. .\(\displaystyle =\, cLab\)

Does this look ok




Next I have this problem:

\(\displaystyle \mbox{11. Find the absolute maximum and minimum values of}\)

\(\displaystyle \mbox{the func}\mbox{tion }\, f(x,\, y)\, =\, (x\, -\, 1)^2\, +\, (y\, +\, 1)^2,\, \mbox{ over}\)

\(\displaystyle \mbox{the region }\, R\, =\, \left\{(x,\, y)\, :\, x^2\, +\, y^2\, \leq\, 4\right\}\)

I'm pretty sure you have to use Lagrange multipliers on this one I know you first need to take partial derivatives of the function then set up the scalar equations involving lamba, I know that much but I'm stuck there.




The third problem I have is this:

\(\displaystyle \mbox{14. Eval}\mbox{uate the inte}\mbox{gral:}\)

\(\displaystyle \displaystyle{\iint\limits_R}\, (y^2\, +\, xy\, -\, 2x^2)\, dA\)

\(\displaystyle \mbox{where }\, R\, \mbox{ is the region bounded by the lines }\, y\, =\, x,\)

\(\displaystyle y\, =\, x\, -\, 3,\, y\, =\, -2x\, +\, 3,\, y\, =\, -2x\, -\, 3\)

For this problem I graphed it out and you can see that there are two sets of parallel lines and the region is rectangular so I think you can use a change of variable. I tried v=y and u=2x+y but it didn't work.
 

Attachments

  • number 5.png
    number 5.png
    1.8 KB · Views: 11
  • 11.jpg
    11.jpg
    6.9 KB · Views: 4
  • number 14.jpg
    number 14.jpg
    8.1 KB · Views: 3
Last edited by a moderator:
Ok the first problem is:

The output Q of an economic system subject to two inputs, such as labor L and capital K, s often modeled by the Cobb-Douglas production function Q(L;K) = cLaKb, where a; b and c are positive real numbers. When a+b = 1, the case is called constant returns to scale. Suppose
a = 1
3 , b = 2
3 and c = 40.

A) If L is held constant at L = 10, write the function that gives the dependence of Q on K.
B) If K is held constant at K = 15, write the function that gives the dependence of Q on L

\(\displaystyle Q(L,\, K)\, =\, cLaKb\)

\(\displaystyle \log Q(L,\,K)\, =\, \log c\, +\, \log L\, +\, \log a\, +\, \log K\, +\, \log b\)

\(\displaystyle \Rightarrow\, \dfrac{1}{Q} \dfrac{\partial Q}{\partial L}\, =\, \dfrac{1}{L}\). . . . .\(\displaystyle \dfrac{1}{Q} \dfrac{\partial Q}{\partial K}\, =\, \dfrac{1}{K}\)

. . . . .\(\displaystyle \dfrac{\partial Q}{\partial L}\, =\, \dfrac{Q}{L}\). . . . . ..\(\displaystyle \dfrac{\partial Q}{\partial K}\, =\, \dfrac{Q}{K}\)

. . . . ..\(\displaystyle Q_L\, =\, \dfrac{Q}{L}\). . . . . . .\(\displaystyle Q_K\, =\, \dfrac{Q}{K}\)

. . . . . . . .\(\displaystyle =\, \dfrac{cLaKb}{L}\). . . . . . .\(\displaystyle =\, \dfrac{cLaKb}{K}\)

. . . . . . . .\(\displaystyle =\, caKb\). . . . . . .. .\(\displaystyle =\, cLab\)

Does this look ok




Next I have this problem:

\(\displaystyle \mbox{11. Find the absolute maximum and minimum values of}\)

\(\displaystyle \mbox{the func}\mbox{tion }\, f(x,\, y)\, =\, (x\, -\, 1)^2\, +\, (y\, +\, 1)^2,\, \mbox{ over}\)

\(\displaystyle \mbox{the region }\, R\, =\, \left\{(x,\, y)\, :\, x^2\, +\, y^2\, \leq\, 4\right\}\)

I'm pretty sure you have to use Lagrange multipliers on this one I know you first need to take partial derivatives of the function then set up the scalar equations involving lamba, I know that much but I'm stuck there.




The third problem I have is this:

\(\displaystyle \mbox{14. Eval}\mbox{uate the inte}\mbox{gral:}\)

\(\displaystyle \displaystyle{\iint\limits_R}\, (y^2\, +\, xy\, -\, 2x^2)\, dA\)

\(\displaystyle \mbox{where }\, R\, \mbox{ is the region bounded by the lines }\, y\, =\, x,\)

\(\displaystyle y\, =\, x\, -\, 3,\, y\, =\, -2x\, +\, 3,\, y\, =\, -2x\, -\, 3\)

For this problem I graphed it out and you can see that there are two sets of parallel lines and the region is rectangular so I think you can use a change of variable. I tried v=y and u=2x+y but it didn't work.
You have posted here before. We prefer that you post one problem in a thread.

A Cobb-Douglas production function NORMALLY looks like \(\displaystyle Q = cL^aK^b,\ not\ Q = cLaKb.\)

So \(\displaystyle Q = cL^aK^b \implies ln(Q) = ln(c) + a * ln(L) + b * ln(K) \implies \dfrac{\delta Q}{\delta L} * \dfrac{1}{Q} = \dfrac{a}{L} \implies \dfrac{\delta Q}{\delta L} = \dfrac{aQ}{L} = acL^{(a - 1)}K^b.\)

Also \(\displaystyle Q = cL^aK^b \implies ln(Q) = ln(c) + a * ln(L) + b * ln(K) \implies \dfrac{\delta Q}{\delta K} * \dfrac{1}{Q} = \dfrac{b}{K} \implies \dfrac{\delta Q}{\delta L} = \dfrac{bQ}{K} = bcL^aK^{(b - 1)}.\)

Please check what equation you are supposed to use for Cobb-Douglas production functions.

Also what in the world does 3, b = 2 mean? Does 3 and c = 40 mean 3 + c = 40 so c = 37?

As for your second problem, I at least would start by finding whether any local maxima and minima exist and, if so, where they exist before worrying about Lagrangian multipliers. Only after that would I bring the multipliers into play. Furthermore, I see the need for only one multiplier when you get to that stage.
 
You have posted here before. We prefer that you post one problem in a thread.

A Cobb-Douglas production function NORMALLY looks like \(\displaystyle Q = cL^aK^b,\ not\ Q = cLaKb.\)

So \(\displaystyle Q = cL^aK^b \implies ln(Q) = ln(c) + a * ln(L) + b * ln(K) \implies \dfrac{\delta Q}{\delta L} * \dfrac{1}{Q} = \dfrac{a}{L} \implies \dfrac{\delta Q}{\delta L} = \dfrac{aQ}{L} = acL^{(a - 1)}K^b.\)

Also \(\displaystyle Q = cL^aK^b \implies ln(Q) = ln(c) + a * ln(L) + b * ln(K) \implies \dfrac{\delta Q}{\delta K} * \dfrac{1}{Q} = \dfrac{b}{K} \implies \dfrac{\delta Q}{\delta L} = \dfrac{bQ}{K} = bcL^aK^{(b - 1)}.\)

Please check what equation you are supposed to use for Cobb-Douglas production functions.

Also what in the world does 3, b = 2 mean? Does 3 and c = 40 mean 3 + c = 40 so c = 37?

As for your second problem, I at least would start by finding whether any local maxima and minima exist and, if so, where they exist before worrying about Lagrangian multipliers. Only after that would I bring the multipliers into play. Furthermore, I see the need for only one multiplier when you get to that stage.

How about the last problem? I've been trying to solve it using a change of variable using the jacobian. I found that u=y-x and v=-2x-y solving for x and y you get x= (u-v)/3 and y = (4u-v)/3 the for my jacobian found dV to be 1/3 is this right ?
 
Top