Answer:
Option C is correct.
Hans will choose some of each commodity but more y than x.
The number of units of x and y that will maximize Hans' utility and fit well with his budget constraint are 0.854 units of x and 1.334 units of y.
Step-by-step explanation:
Hans wants to maximize utility but has a cost constraint. It seems like this is a 'finding the maximum of a multivariable function' problem.
U(x,y) = 5x² + 2y²
The constraint can be given as
Sum of the Cost of the number units of each commodity he will get must not exceed $27
Price of one commodity x = $16
Price of one commodity y = $10
16x + 10y ≤ 27
But since fractional units are possible,
16x + 10y = 27.
So, we want to maximize U(x,y) = 5x² + 2y² subject to the function of 16x + 10y = 27.
There are a number of ways to go About this, but the best of them all is the Lagrange function method.
We will write a Lagrange function,
L(x,y) = U(x,y) - λ(constraint function)
Constraint function = 16x + 10y - 27
L(x,y) = 5x² + 2y² - λ(16x + 10y - 27)
where λ can be a function of x and y
So, we take the derivatives of L with respect to x, y and λ. The values of x and y at this point are our answers.
L(x,y) = 5x² + 2y² - λ(16x + 10y - 27)
(∂L/∂x) = 10x - 16λ
At the maximum value we require, (∂L/∂x) = (∂L/∂y) = (∂L/∂λ) = 0
10x - 16λ = 0
λ = 10x/16 = 5x/8
(∂L/∂y) = 4y - 10λ
4y - 10λ = 0
λ = 4y/10 = 2y/5
(∂L/∂λ) = 16x + 10y - 27
16x + 10y - 27 = 0
Equating the values of λ obtained from the first two partial derivatives
λ = (5x/8) = (2y/5)
x = 16y/25
Putting this value for x in the third partial derivatives or the constraint function
16x + 10y - 27 = 0
16(16y/25) + 10y = 27
(256y/25) + 10y = 27
506y/25 = 27
506y = 27×25
y = 27×25/506 = 675/506 = 1.334 units.
x = 16y/23 = 0.854 units.
So, the number of units of x and y that will maximize Hans' utility and fit well with his budget constraint are 0.854 units of x and 1.334 units of y.
