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Submitted By josephk

Words 1054

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Words 1054

Pages 5

Q=K^(1/4) √L

Solution

Min: C_((K,L))=r_K .k+ w_L .l=20k+10l

Constraint: Q=k^(1/4) √l

Using Lagrangian method

L_((k,l,λ))=20k+10l- λ(k^(1/4) .l^(1/2)-Q)

dL/dk=20- 1/4 λk^(-3/4) .l^(1/2)=0 dL/dl=10- 1/2 λk^(1/4) .l^(-1/2)=0 dL/dλ=- 1/4 λk^(1/4) .l^(1/2)-Q=0

Equating

20- 1/4 λk^(-3/4) .l^(1/2)= 0 20= 1/4 λk^(-3/4) .l^(1/2) λ= (80k^(3/4))/l^(1/2) 10- 1/2 λk^(1/4) .l^(-1/2)=0 10= 1/2 λk^(1/4) .l^(-1/2) λ= (20l^(1/2))/k^(1/4)

(80k^(3/4))/l^(1/2) = (20l^(1/2))/k^(1/4)

80k=20l

4k=l

Substituting Lagrangian relation into Output equation

Q=K^(1/4) √L

Q=k^(1/4) 〖(4k)〗^(1/2)

Q=k^(1/4) 〖2k〗^(1/2)

Q/2=k^(3/4)

k= Q^(4/3)/2^(4/3) = Q^(4/3)/∛16= Q^(4/3)/(2∛2)

Solving for l

4k=l

2^2 (Q^(4/3)/2^(4/3) )=l

l= Q^(4/3)/2^(2/3) = Q^(4/3)/∛4

Minimum total expenditure on capital and labour in terms of Q

C_((K,L))=20k+10l

C_((K,L) )=2^2.5.(Q^(4/3)/2^(4/3) )+10(Q^(4/3)/2^(2/3) )

C_((K,L) )= (5Q^(4/3)+10Q^(4/3))/2^(2/3) = 〖15Q〗^(4/3)/2^(2/3) = 〖2^2.5.Q〗^(4/3)/2^(2/3) = (5Q^(4/3))/∛2

Information x denotes units of good X y denotes units of good Y

Cx denotes the unit cost of good X in dollars = $1

Cy denotes the unit cost of good Y in dollars = $1

0 < β < 1

M denotes the maximum amount of dollars to spend in the two goods Utility Function

U_((x,y))=〖(x^β+ 3y^β)〗^(1/β)

Solution Max: U_((x,y))=〖(x^β+ 3y^β)〗^(1/β)

Constraint: M_((x,y))= C_x x+ C_y y

Using Lagrangian method

L_((x,y,λ) )= (x^β+ 3y^β )^(1/β)- λ(x+y-M)

dL/dx= 1/β (x^β+ 3y^β )^(1/β-1) .(βx^(β-1) )- λ=0

dL/dy= 1/β (x^β+ 3y^β )^(1/β-1) .(3βy^(β-1) )- λ=0

dL/dλ= -x-y+M=0…...

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