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economic mathematic questions. need specific process!
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Explanation & Answer

Hi, Please check the attached file for details, let me know if you have any questions, thank you. James
Set1
1. 2000= 20K3/5L2/5
100= K3/5L2/5
1005 = k3L2
Solve for L
L = 105/k3/2
The object function is
C = 1000k+600L = 1000K + 600*105/k3/2
dC/dK = 1000+6*107 (-3/2)/k5/2
let dC/dK=0
1000 – 9*107/K5/2 = 0
K =(90000)2/5
= (300)4/5
= 95.873
L = 105/K3/2
= 105/(300)6/5
=1/3 * 103/(300)1/5
= 106.526
The minimum cost is
C = 1000K+600L = 95873+ 63915
=159788
To prove that the answer is indeed absolute minimum, we take second derivative of C
with respect K,
d2C/dK2 = -9*107 *(-5/2)/k7/2
=9/4 *108/k7/2>0
Thus it is a minimum. Since C = 1000k+600*105/k3/2, K is in (0,), as K approach infinity,
so does C and as K 0 , C also approach infinity. In addition, since C only has one
critical point, and it is a relative minimum, it must be a absolute minimum.
2. a) The revenue is
R = Pq = (54-0.4q)q
The profit is
Pr = R-C = (54-04q)q-10q-80
= 44q-0.4q2-80
dPr/dq = 44-0.8q =0
the daily output for maximum profit is
q = 55
the price at maximum profit is
P = 54-0.4q = 54-22 = 32
and the maximum profit is
Pr = 44*55-0.4*552 -80
=1130
To prove it is absolute maximum, we take second derivative
d2pr/dq2 = -0.8,
the profit function is a parabola open downward, thus the critical point we found above
is the absolute maximum.
b) Now the profit function is
Pr = = 44q-0.4q2-80 – 1.6q
=42.4q-0.4q2 -80,
The price maximize the profit is found by the equation
dpr/dq = 0
42.4-0.8q =0
q= 53
P = 54-0.4q =$32.8
Thus the firm should pass $0.8 to the consumer.
3. a
Let the derivative =0, we have
-4x2 – 6x +4 =0
2x2 +3x -2 =0
(2x-1)(x+2) = 0
Th...
