Minimize the following cost function ; TC = 1/3q^3 - 8 . 5q^2 + 60q+27
TC = q3/3 - 8 .5q2 + 60q + 27
dTC/dq = 3q2/3 - 17q + 60
= q2 - 17q + 60
put dTC/dq = 0
q2 - 17q + 60 = 0
q2 - 5q - 12q + 60 = 0
q(q - 5) -12(q - 5) = 0
(q - 12)(q - 5) = 0
q - 12 = 0 or q - 5 = 0
q = 12 or q = 5
SOC
d2TC/dq2 = 2q - 17
at q = 5
d2TC/dq2 = 2(5) - 17 = 10 - 17 = - 7
d2TC/dq2 < 0
Therefore cost is maximum at q = 5
at q = 12
d2TC/dq2 = 2(12) - 17 = 24 - 17 = 7
d2TC/dq2 > 0
so cost is minimum at q = 12
Minimumm TC = (123)/3 - 8.5(12)2 + 60(12) + 27
= 576 - 1224 + 720 + 27
= 1323 - 1224
= 99
Get Answers For Free
Most questions answered within 1 hours.