Hans has $27 which he decides to spend on x and y. Commodity x costs $16 per unit and commodity y costs $10 per unit. He has the utility function U(x, y)- 5x2+2y2 and he can purchase fractional units of x and y. Hans will choose:a. only X b. only yc. some of each commodity but more y than x. d. some of each commodity but more x than y.

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.