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SECOND SHORT TEST ON INCREMENT APPROXIMATION
Consider an utility function U(x,y)=Jx^3 + y
5 x N
1 ê 5 where x denotes comsuption of A and y denotes
comsumption of B. Current situation is (x,y)= ( 8 , 6 ), and current variations are (Dx,Dy)= (0.15%,-0.13(units)).
(a) Calculate the expression of each partial derivative.
∂U ∂x =^
3 x^2 - y x^52 5 Kx^3 + y x^5 O 4 ë 5
∂U ∂y =^
y^4 x Kx^3 + y x^5 O 4 ë 5
(b) Calculate the value of each partial derivative at current situation (x,y)= ( 8 , 6 ).
∂U ∂x H8, 6L^ =^ 0. ∂U ∂y H8,6L^ =^ 0.
(c) Write the formula for approximating the following partial increments:
(c1) DUx( 8 , 6 )(0.15%) ª ∂ ∂Ux H8, 6LâH0.012L
(c2) DUy( 8 , 6 )(-0.13(units)) ª ∂ ∂Uy H8, 6LâH-0.13L
(d) Obten the numerical results of the formulas in sections (c1) and (c2)
HdL DUxH8,6LH0.012Lª ∂ ∂Ux H8,6LâH0.012L = 0.0409329âH0.012L = 0.
DUyH8,6LH-0.13Lª ∂ ∂Uy H8,6LâH-0.13L = 0.470292âH-0.13L =-0.
(e) Write the formula for approximating the following total increment: DU( 8 , 6 )(0.15%,-0.13(- units)).
HeL DUH8,6LH0.012,-0.13L ª ∂ ∂Ux H8,6LâH0.012L + ∂ ∂Uy H8,6LâH-0.13L
(f) Obtain the numerical value of the formula in section (e)
HfL DUH8,6LH0.012,- 0.13L ª
∂ U ∂ x
H8,6LâH0.012L +
∂ U ∂ y
H8,6LâH- 0.13L = - 0.
(g) Calculate the exact value of those partial increments in sections (c1) and (c2).
HgLValor exacto de Hc1L DUxH8,6LH 0.012L = UH8.012, 6L - UH8,6L = 0.
Valor exacto de Hc2L DUyH8,6LH-0.13L= UH8, 5.87L - UH8,6L =-0.
(h) Calculate the error of the approximations obtained in sections (c1) and (c1) as compared with the exact value in section (g).
HhLError of Hc1L: Errorx = -0.63962%
Error of Hc2L: Errory = 1.55002%
(i) Calculate the exact value of the total increment in section (e).
HiL DUH8,6LH0.012,-0.13L= UH8.012, 5.87L- UH8,6L = -0.
(j) Calculate the error of the approximations obtained in section (e) as compared with the exact value in section (i).
HjL Error = 1.77496%
(k) Considering simultaneous variation (Dx,Dy)= (0.15%,-0.13(units)), approximate the new value of demand. HkL UH8.012, 5.87L = 4.24852 ª UH8, L +
∂ U ∂ x
H8,6LâH0.012L +
∂ U ∂ y
H8,6LâH- 0.13L = 4.
2 Short Test Increment Approximation 2. Solved .nb
