12.1

1. (a) Let x and v denote total production in industries A and I, respectively. Then * = \x + {y + 60 and >■ = |x + ¿y + 60. So fx - \y = 60 and -¿* + !>- = 60. (b) x = 320/3'and y= 1040/9.

3. 0.8x1 - 0.3x2 = 120 and -0.4x, + 0.9xo = 90. with solution x\ = 225 and x2 = 200.

12.2

!. a+b= (3).-* = (l^),2a + 3b= (J^,and-5a+2b= 3. a\ =0, a2 = 1/3, and = 1.

5. (a) x, = 0 for all i. (b) Nothing, because 0 • x = 0 for all x.

9. (a) (i) Possible, with 9 = 1/2. (ii) Impossible, (iii) Impossible.

(b) (i) Proportion of lead 8 = 1/2. (ii) If output can be thrown away, the proportion of lead can be 6 = 2/3. (iii) Impossible in any case, because (1 _ 0) 4 + $ 3 < 9 for all 9 e [0,1].

12.3

3. (a)jc, = 2,x2 = -l. (b) Suppose x, (1.2. l)+x2(-3. 0,-2) = (-3.6, 1). Then x\ — 3x2 = -3, 2x\ = 6, and X) — lx2 = 1. The first two equations yield x\ = 3 and x2 = 2; then the last equation is not satisfied.

5. (a) A straight line through (0, 2, 3) parallel to the x-axis.

(b) A plane parallel to the z-axis through the line y = x in the xy-plane.

12.4

1. a-a = 5, a-b = 2, and a-(a + b) = 7. We see that a-a + a-b = a - (a + b). 3. The pairs of vectors in (a) and (c) are orthogonal.

5. The vectors are orthogonal iff their scalar product is 0, that is iff x2 - x -

% - 2x + x = x2 -2x = 0. which is the case for x = -2 and x = 4. 7. Use the rules in [12.14].

9. (a) The firm's revenue is p • z. Its costs are p • x. (b) Profit = revenue -

costs = p • z - p. x = p • (z — x) = p • y. If p • y < 0, the firm makes a loss equal to —p - y.

11. (a) Input vector = ^ (b) 0utput vector = ^ (c) Cost= (1, 3) ^ =3

(d) Revenue = (l. 3) ^ = 2 (e) VaJue Qf net outpm = (L 3) ^ = 2 ~ 3 = -1 (0 Loss = cost - revenue = 3-2=1.

1. (a) *, = 3 + It, *2 = -2 + 4r, and *3 = 2 - r (b) *, = 1, *2 = 3 - r, and x3 = 2 + r

3. - 3*2 - 2*3 = -3. (One method: (5. 2, 1) - (3,4. -3) = (2, -2,4) and (2, -1,4) - (3,4, —3) = (— 1, -5, 7) are two vectors in the plane. The normal (pi, pi- Pj) must be orthogonal to both these vectors, so (2, -2,4) • (PuPi-Ps) = 2p\ -2p1 + Apl = 0 and (-1. -5,7) ■ (pi, pi, p3) = —p\ - 5p2 + 7/?3 = 0. One solution to these two equations is (p\. pi, p3) — (1, -3, -2). Then using formula [12.23] with (a,. a2. a3) = (2, -1,4) yields (1,-3, -2) - (*i - 2, *2 + 1, *3 -4) = 0, which reduces to *] - 3*-? - 2*3 = -3-)

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