Answer to Question #218087 in Microeconomics for Yves10-Official

Question

You are given the following utility function and price of commodities q1 and q2:

U = 3q1+q1q2-5q2-15

P1=3 and p2=2

If the corresponding bugdet is 20.

i. Write the consumer's budget equation

ii.construct a constrained utility maximization problem out of the information given above

iii. Write the augmented objective function?

Iv. Find the optimum level of U

V. Is the second order condition for a maximum satisfied?

Vi. Find the levels of q1 and q2 that will satisfy the first order condition for a maximum.

Answer 1

Given

U = 3q1+q1q2-5q2-15

P1=3 and p2=2

Utility is maximized at the point where indifference curves are tangential to the budget constraint .  

i) Budget Constraint : 

There are two sides to BC expenditure side and the income side . 

Expenditure side $= P_1q_1 + P_2q_2$

Income Side$= m = 20$

Budget Constraint :

$P_1q_1 + P_2q_2 =< 20$

ii) Utility maximization problem is optimized where : MRS ( marginal rate of substitution )$= \frac{P_1}{P_2}$

iii) Objective function is referred to the function which is to be optimized subject to the constraint .

Here utility function is to be maximized :

$U_{max} = 3q_1+q_1q_2-5q_2-15$

$s.t P_1q_1 + P_2q_2 =< 20$

iv) Optimization point is where :

$MRS = \frac{P_1}{P_2}\\ MRS =\frac{ MU_{q1}}{MU_{q2}} \\ MU_{q1} = u'(q_1) = 3 + q_2\\ MU_{q2} = u'(q_2) = q_1 - 5$

v)

Second Order Condition of the utility maximization problem is that the marginal utility of of any good should be diminishing .

$U''(q_1)$ should less than or equal to 0. 

$U''(q_1 ) = \frac{du^2}{(dq_1)^2} = \frac{d(3 + q_2)}{dq_1} = 0$

Hence SOC is justified .

vi

Putting in optimization equation :

$\frac{(3 + q_2)}{(q_1 - 5 )} = \frac{P_1}{ P_2}\\ \frac{(3 + q_2)}{(q_1 - 5 )} = \frac{ 3}{2}\\ 6 + 2q_2 = 3q_1 - 15 \\ 3q_1 - 2q_2 = 21 --------(i) also\space budget\space constraint\space is \space given \space as :\\ P_1q_1 + P_2q_2 = 20 \\ 3q_1 + 2q_2 = 20 --------(ii)\\ Solving\space equation\space (i) \space and\space (ii)\space simultaneously\space using\space elimination\space we\space get :\\ 6q_1 = 41 \\ q_1 = \frac{41}{6 }= 6.83\\ Putting\space this\space into\space (i)\\ q_2 = 2.44$