Question

for the constrained optimisation problem below, find the demand functions and the lagrangian multiplier. interpret the multiplier. max U(x, y)=4x ² + 3y ² subject to 3x + 4y=100

Answer 1

Given,

$max U(x,y)=4x^2+3y^2 \space subject\space to \space 3x+4y=100$

The Lagrangian function for the above optimization problem is given below:

$L=4x^2+3y^2+λ(100−3x−4y)$

Let us find the first order condition

$\frac{∂L}{∂x}=8x−3λ\\\frac{∂L}{∂y}=6y−4λ\\\frac{∂L}{∂λ}=100−3x−4y$

Let us substitute all the first order conditions equal to zero

$\frac{∂L}{∂x}=0\\8x−3λ=0\\8x=3λ.....equation \space1\\\frac{∂L}{∂y}=0\\6y−4λ=0\\6y=4λ...equation \space 2\\And,\\\frac{∂L}{∂λ}=0\\100−3x−4y=0\\100=3x+4y.....equation\space 3$

Let us divide equation 1 by 2

$\frac{8x}{6y}=\frac{3λ}{4λ}\\x=\frac{3×6y}{4×8}\\x=\frac{9y}{16}....equation\space 4$

Let us substitute the value of x in equation 3

$100=3×\frac{9y}{16}+4y\\100=\frac{27y}{16}+4y\\100=\frac{27y+64y}{16}\\100=\frac{91y}{16}\\y=\frac{100×16}{91}\\y=17.58$

Let us substitute the value of y=17.58 in equation 4

$x=\frac{9×17.58}{16}\\x=9.89$

Let us substitute the values of x in equation 1 for the value of the multiplier

$8×9.89=3λ\\λ=\frac{8×9.89}{3}\\λ=26.37 \space approx$

The value of the Lagrangian multiplier means if the budget of the consumer increases by 1 unit then the total utility will increase by 26.37 units.

 

Note: in the above part the exact demand has been calculated because the prices and budget were given in the budget constraint. To find the demand function let us use the budget constraint as given below:

$Budget\space constraint: xp_x+yp_y=m\\L=4x^2+3y^2+λ(m−xp_x−yp_y)$

Let us find the first order conditions

$\frac{∂L}{∂x}=8x−λp_x\\\frac{∂L}{∂y}=6y−λp_y\\\frac{∂L}{∂λ}=m−xp_x−yp_y$

Let us substitute all the first order conditions equal to zero

$\frac{∂L}{∂x}=0\\8x−λp_x=0\\8x=λp_x.....equation \space 1\\\frac{∂L}{∂y}=0\\6y−λp_y=0\\6y=λp_y...equation\space 2\\And,\\\frac{∂L}{∂λ}=0\\m−xp_x−yp_y=0\\m=xp_x+yp_y.....equation\space 3\\ divide \space equation\space 1 \space by \space 2\\\frac{8x}{6y}=\frac{λp_x}{λp_y}\\x=\frac{6yp_x}{8p_y}....equation \space 4$

Let us substitute the value of x in equation 3

$m=px×\frac{6yp_x}{8p_y}+yp_y\\m=\frac{6yp^2_x}{8p_y}+yp_y\\m=\frac{6yp^2_x+8yp^2y}{8p_y}\\m=y(\frac{6p^2_x+8p^2_y}{8p_y})\\y=\frac{m}{\frac{6p^2_x+8p^2_y}{8p_y}}\\y=\frac{8p_ym}{6p^2_x+8p^2_y}...(demand\space function \space of \space good \space y)$

Let us substitute the value of y in equation 4

$x=\frac{6×\frac{8p_ym}{6p^2_x+8p^2_y}×px}{8py}\\x=\frac{48mp_xp_y}{(6p^2_x+8p^2_y)8p_y}\\x=\frac{48mp_xp_y}{48p^2_xp_y+64p^3y}..(demand \space function\space of \space good\space x)$

The demand equations as a function of px, py, and m are given below:

$x=\frac{48mp_xp_y}{48p^2_xp_y+64p^3_y}\\y=\frac{8mp_y}{6p^2_x+8p^2_y}$