Utility Maximization
This scene finds the best bundle where the budget line meets an indifference curve.
The optimum is where the budget line just touches the curve. Move prices or wealth and the optimum slides along. At that spot the MRS always equals the price ratio, as in the formula below. You can check it numerically at every moment.
For example, you have 💰 6,000 won for 🍎 apples and 🍌 bananas, each 1,000 won. If you like both equally, three apples and three bananas are best. If apples get pricier, you buy fewer apples and more bananas.
max x u ( x ) s.t. p ⋅ x ≤ w \max_x u(x) \quad \text{s.t.} \quad p \cdot x \le w x max u ( x ) s.t. p ⋅ x ≤ w x ∗ ( p , w ) = ( a w p 1 , ( 1 − a ) w p 2 ) x^*(p, w) = \left(\frac{aw}{p_1}, \frac{(1-a)w}{p_2}\right) x ∗ ( p , w ) = ( p 1 a w , p 2 ( 1 − a ) w ) Part of these learning paths
Examples & Exercises
With u = x 1 1 / 2 x 2 1 / 2 u = x_1^{1/2}x_2^{1/2} u = x 1 1/2 x 2 1/2 p 1 = p 2 = 1 p_1 = p_2 = 1 p 1 = p 2 = 1 w = 60 w = 60 w = 60
L = x 1 1 / 2 x 2 1 / 2 + λ ( w − p 1 x 1 − p 2 x 2 ) \mathcal{L} = x_1^{1/2}x_2^{1/2} + \lambda(w - p_1 x_1 - p_2 x_2) L = x 1 1/2 x 2 1/2 + λ ( w − p 1 x 1 − p 2 x 2 ) 1 2 x 1 − 1 / 2 x 2 1 / 2 = λ p 1 , 1 2 x 1 1 / 2 x 2 − 1 / 2 = λ p 2 \frac{1}{2}x_1^{-1/2}x_2^{1/2} = \lambda p_1, \quad \frac{1}{2}x_1^{1/2}x_2^{-1/2} = \lambda p_2 2 1 x 1 − 1/2 x 2 1/2 = λ p 1 , 2 1 x 1 1/2 x 2 − 1/2 = λ p 2 M R S = p 1 / p 2 MRS = p_1/p_2 M R S = p 1 / p 2 a = 1 / 2 a = 1/2 a = 1/2 p = ( 1 , 1 ) p = (1,1) p = ( 1 , 1 ) w = 60 w = 60 w = 60 Show proof
Dividing the two first-order conditions eliminates λ \lambda λ x 2 x 1 = p 1 p 2 \frac{x_2}{x_1} = \frac{p_1}{p_2} x 1 x 2 = p 2 p 1 M R S = p 1 / p 2 MRS = p_1/p_2 M R S = p 1 / p 2
Since p 1 = p 2 p_1 = p_2 p 1 = p 2 x 1 = x 2 x_1 = x_2 x 1 = x 2 x 1 + x 2 = 60 x_1 + x_2 = 60 x 1 + x 2 = 60 x ∗ = ( 30 , 30 ) x^* = (30, 30) x ∗ = ( 30 , 30 ) ( 30 , 30 ) (30,30) ( 30 , 30 )
Apply Roy's identity, the formula below, to Cobb-Douglas indirect utility v ( p , w ) v(p, w) v ( p , w )
v ( p , w ) = a a ( 1 − a ) 1 − a w p 1 a p 2 1 − a v(p, w) = a^a (1-a)^{1-a} \frac{w}{p_1^a p_2^{1-a}} v ( p , w ) = a a ( 1 − a ) 1 − a p 1 a p 2 1 − a w x 1 = − ∂ v / ∂ p 1 ∂ v / ∂ w x_1 = -\frac{\partial v/\partial p_1}{\partial v/\partial w} x 1 = − ∂ v / ∂ w ∂ v / ∂ p 1 Raised p 1 = 2 p_1 = 2 p 1 = 2 x 1 = 15 x_1 = 15 x 1 = 15 Show proof
Differentiating v v v ∂ v / ∂ p 1 = − a ⋅ v / p 1 \partial v/\partial p_1 = -a \cdot v/p_1 ∂ v / ∂ p 1 = − a ⋅ v / p 1 ∂ v / ∂ w = v / w \partial v/\partial w = v/w ∂ v / ∂ w = v / w
Hence − ∂ v / ∂ p 1 ∂ v / ∂ w = a v / p 1 v / w = a w p 1 -\frac{\partial v/\partial p_1}{\partial v/\partial w} = \frac{a v/p_1}{v/w} = \frac{aw}{p_1} − ∂ v / ∂ w ∂ v / ∂ p 1 = v / w a v / p 1 = p 1 a w x 1 = 0.5 ⋅ 60 / 2 = 15 x_1 = 0.5 \cdot 60/2 = 15 x 1 = 0.5 ⋅ 60/2 = 15
4. Classical Demand Theory