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Unformatted text preview: 15.2 SOLUTIONS 1057 30. (a) Let t be the number of years since 1960 and let P(t) be the population in millions in the year 1960 + t. We assume that P = Ceat, and therefore InP = at + InC. So, we plot In P agaist t and find the line of best fit. Our data points are (0, In 180), (10, In 206), and (20, In 226). Applying the method ofleast squares to find the bestfittng lie, we find that a = In 226 ~ In 180 r: 0.0114, In C = In ~06 _ In ~26 + 5In6180 r: 5.20 Then, C = è.20 = 181.3 and so P(t) = 181.3eo.OIl4. In 1990, we have t = 30 and the predicted population in millons is P(30) = 181.3eo.01141(30) = 255.3. (b) The diference between the actual and the predicted population is about 6 millon or 2 ~ %. Given that only thee data points were used to calculate a and c, ths discrepancy is not surrising. Thus, the 1990 census data does not mean that the assumption of exponential growt is unjustified. (c) In 2010, we have t = 50 and P(50) = 320.7. 31. Let P(K, L) be the profit obtaed using K units of capita and L units of labor. The cost of production is given by C(K, L) = kK + PL, and the revenue fuction is given by R(K, L) = pQ = pAKa Lb. Hence, the profit is P = R  C = pAKa L b  (kK + PL). In order to find local maxima of P, we calculate the paral derivatives and see where they are zero. We have: ôP = AKa1 Lb _ k ôK ap , ~~ = bpAKa LbI .e. The critical points of the function P(K, L) are solutions (K, L) of the simultaeous equations: ~ =pAKaILb, a ~ = pAKaLbl. Multiplying the fist equation by K and the second by L, we get kK PL;  b' and so .ea K = kb L. Substitutig for K in the equation k / a = pAKa I L b, we get: ~ =pA (~~rI LaILb. We must therefore have Liab = pA G~) a (Ð aI Hence, if a + b i 1, _ ( (a) a (.e) (ai)) I/(Iab) . L  pA k b ' 1056 Chapter Fifteen ¡SOLUTIONS 15.2 SOLUTIONS 1057 'fi' Since W, h i 0, we can divide the first equation by the second giving 2wh2 2048 2hw2 = 1024' thus ~ =2, W 30. (a) Let t be the number of years since 1960 and let P(t) be the population in millons in the year 1960 + t. We assume that P = Ceat, and therefore InP = at + InC. So, we plot In P agaist t and find the line of best fit. Our data points are (0, In 180), (10, In 206), and (20, In 226). Applying the method of least squares to find the bestfittg lie, we find that a = In 226 ~ In 180 r: 0.0114, In C = In 206 _ In 226 + 5 In 180 r: 5 20 3 6 6 . so h=2w. Substituting ths in Ch = 0, we obtain h3 = 2048, so h = 12.7 cm. Thus w = h/2 = 6.35 cm, and L = 512/(wh) == 6.35 cm. Now we check that these dimensions minimize the cost C. We find that 2 4096 2048 2 D = ChhCww  Chw = ()( w3 )  2 , and at h = 12.7, w = 6.35, Chh )0 0 and D = 16  4 )0 0, thus C has a local minimum at h = 12.7 and w = 6.35. Since C increases without bound as w, h l 0 or 00, ths local mium must be a global mium. Therefore, the dimensions of the box that minimize the cost are w = 6.35 cm, L = 6.35 cm and h = 12.7 cm. 28. The square of the distace from the point (x, y, z) to the origin is28....
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This note was uploaded on 10/20/2009 for the course MATH 254 taught by Professor Hellin during the Fall '09 term at Oregon State.
 Fall '09
 Hellin

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