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# D now tr tc 8 q 12 025 q when d dq 4 q 12 025 0

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(d) Now = TR – TC = 8 Q 1/2 – 0.25 Q . When d / dQ = 4 Q –1/2 – 0.25 = 0, profit-max- imizing Q ( Q* ) = (4/0.25) 2 = 256 boxes/period. P = 8 Q –1/2 = 8(256 –1/2 ) = \$0.50/box, and = TR – TC = PQ – 0.25 Q = (0.5)(256) – 0.25(256) = \$64/period. [Again, we know that Q* = 64 is a point of maximum rather than minimum prof- it, because the second derivative d 2 / dQ 2 = –2 Q –3/2 is negative for all Q > 0.] 5. (a) The budget constraint is Y = 60 – 2 X , and so U may be rewritten U = X 2 Y = X 2 (60 – 2 X ) = 60 X 2 – 2 X 3 . Hence dU / dX = 120 X – 6 X 2 , and setting d U / dX = 0, X = 20 units, whence Y = 60 – 2 X = 20 units as well. (b) Now the budget constraint is Y = 60 – X . Hence U = X 2 Y may be rewritten U = X 2 (60 – X ) = 60 X 2 X 3 . Now dU / dX =120 X –3 X 2 , and setting d U / dX = 0, we have X = 40 units, whence Y = 60 – X = 20 units, as before. M10-2 MATH MODULE 10: SOLUTIONS TO EXERCISES
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