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The second derivative equals zero at the origin which

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The second derivative equals zero at the origin, which is an inflection point. [This function is depicted in Figure M.10-3(b).] (f) dy/dx = 4/(1 + x ) 2 , and d 2 y / dx 2 = –8/(1 + x ) 3 . The function is not defined at x = –1, and dy/dx nowhere equals zero, so the function has no local maxima or min- ima. Since d 2 y / dx 2 nowhere equals 0, the function has no inflection point. 2. (a) When X = 16 units, maximum Y = 100 – .25 (16) 2 = 36 units, and the slope of the PPF = dY/dX = –0.5 X = –8. (b) When dY/dX = –2 = –0.5 X , we have X = 4 units, and Y = 100 – .25 (4) 2 = 36 units. M ATH M ODULE S olutions to Exercises 10
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3. The maximum for f ( x ) occurs with f ’( x ) = 32 – x = 0, or x = 32, at which point, y = 32 x – .5(32) 2 = 512. The maximum for h ( x ) = (10 + 24 x – .75 x 2 occurs with h ’( x ) = 24 – 1.5 x = 0, or x = 16, at which point, h ( x ) = (10 + 24(16) – .75(16 2 ) = 182. [Note that this is basically the structure of a profit-maximization problem, with f ( x )as the Total Revenue function, g ( x )as the Total Cost function, and h ( x ) as the Profit func- tion.] 4. (a) P = (64 Q –1 ) 1/2 = 8 Q –1/2 . (b) (i) TR = PQ = P (64 P –2 ) = 64 P –1 ; (ii) TR = PQ = (8 Q –1/2 )( Q ) = 8 Q 1/2. (c) =TR – TC = 8 Q 1/2 – 0.5 Q . When d / dQ = 4 Q –1/2 – 0.5 = 0, profit-maximizing Q ( Q *) =(4/0.5) 2 = 64 boxes/period. P = 8 Q –1/2 = 8(64 –1/2 ) = $1/box, and = TR – TC = PQ – 0.5 Q = (1)(64) – 0.5(64) = $32/period. [We know that Q* = 64 is a point of maximum profit rather than minimum profit, because the second deriv- ative d 2 / dQ 2 = –2 Q –3/2 , which is negative for Q > 0.]
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