how will u treat the following while estimating the natinonal income of india reaso a. payment of interest on borrowings by the general gvrmnt b. incrase in price of the shares of a company c. gvrmnt expenditure on sanitation d. growing vegetable in a kitchen garden of the house
Answers
Answer:
Explanation:
1. Differentiate the following functions:
(a) f(x) = 6
+
− + .
(b) () = √
(c) () =
−
( − ).
(d) () =
/(
− )
(e) () = (
−
).
ANS: The derivatives are:
(a)
′
() = 18
2 + 4 − 1.
(b)
′
() = 2⁄√.
(c)
′
() = −
−
( − 2) +
− =
−
(3 − ).
(d)
′
() = [12
2
(2
2 − ) − (4 − 1)4
3
/(2
2 − )
2
.
(e)
′
() = [1
−3
][−3
−4 ⁄ ] = −3 /.
2. Determine whether the following functions are strictly convex, strictly concave, or neither
over the specified intervals:
(a) () =
− + , for x = any real number.
(b) () = , for x > 0.
(c) () =
, for x ≦ .
(d) () =
−
+ , for x ≧ 0.
ANS: The answers are determined by the signs of the second derivatives:
(a)
′′() = 2 > 0, (x) is strictly convex.
(b)
′′() = −1 /
2 < 0, and f(x) is strictly concave.
(c)
′′() =
2
> 0, () is strictly convex.
(d)
′′() = 6 − 4 which does not have a unique sign for x ≥ 0, and f(x) is neither strictly convex
nor strictly concave over the entire interval.
3. Find the values of x1 and x2 which maximize
(,
) = + + − .
−
ANS: Setting the partial derivatives equal to zero,
1 = 5 + 2 − 1 = 0 2 = 10 − 62 + 1 = 0
ℎ equations have the solution x1 = 8, x2 = 3. The second-order conditions for a maximum are
satisfied by this solution:
11 = −1 < 0 |
11 12
21 22
| = |
−1 1
1 −6
| = 5 > 0
2
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4. Let f(x1, x2) = A
, where A, , > 0, be defined for the domain x1, x2 > 0. Demonstrate
that the function is strictly concave within its domain if and only if + < 1.
ANS: If + < 1, the principal minors of the Hessian will alternate in sign, beginning with minus, as
required for strict concavity:
11 = ( − 1)1
−22
< 0
|
11 12
21 22
| = |
( − 1)1
−22
1
−12
−1
1
−12
−1
( − 1)1
2
−2
| =(1 − − )
21
2(−1)
2
2(−1) > 0
Conversely, + ≥ 1 will violate the requirement that the Hessian be positive, and concavity cannot
hold.
5. Find the values for x1 and x2 that maximize f (x1, x2) =
subject to the requirement that
5x1 + 2x2 = 300. Demonstrate that the appropriate second-order condition is satisfied.
ANS: Form the Lagrange function
= 1
22 + (51 + 22 − 300)
Where λ is an undetermined multiplier, and set its partial derivatives equal to zero:
1
= 212 + 5 = 0
2
= 1
2 + 2 = 0
= 51 + 22 − 300 = 0
Substitute 22 = 5 1⁄2 from the first two equations into the third:
5x1 + 51
2
− 300 = 0
Which gives the solution x1 = 40, x2 = 50.
The second-order condition, which requires that the bordered Hessian be positive, is satisfied:
|
22 21 5
21 0 2
5 2 0
| = 401 − 82 = 1200 > 0
6. Find functions of two variables with the domains x1, x2 > 0 that are
(a) Quasi-concave, but not strictly quasi-concave and not concave.
(b) Strictly quasi-concave, but not concave.
(c) Quasi-concave, but not strictly quasi-concave and not strictly concave.
(d) Strictly quasi-concave and concave, but not strictly concave.
ANS: A function is concave if 11 ≤ 0 and H = 1122 − 12
2 ≥ 0, and strictly concave if the strict
inequalities hold. A function is quasi-concave if D = 12
2 12 − 112
2 − 221
2 ≥ 0, and strictly quasi-
3
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concave if the strict inequality holds. The reader may verify that the following functions have the desired
properties by evaluating the appropriate determinants:
(a) (1, 2
) = −(ln 1 − ln 2).
(b) (1, 2
) = 12.
(c) (1, 2
) = 1 + 2.
(d) (1, 2
) = 1
0.52
0.5
.
7. The locus of points of tangency between income lines and indifference curves for given prices
p1, p2 and a changing value of income is called an income expansion line or Engel curve. Show
that the Engel curve is a straight line if the utility function is given by
U =
, > .
ANS: Form the function V = 1
2 + ( − 11 − 22) and set its partial derivatives = 0.
1
−1
2 − 1 = 0 1
− 2 = 0 y - 11 − 22 = 0
Which yields 11 = 22, a positively sloped straight line through the origin.
8. Let a consumer’s utility function be U =
+ . + and his budget constraint
3 + = . Show that his optimum commodity bundle is the same as in Exercise 2.3.
Why is this the case?
ANS:- V is a monotonic transformation of the utility function. Specifically, V = U4 + ln U. bye