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Unformatted text preview: 882 Chapter Twelve SOLUTIONS II ! II :/ (c) Shi the diagram 2 units to the right and 2 units up. See Figue 12.88. y y 021012 x Figure 12.88: f(x  2, y  2) Figure 12.89: f( x, y) (d) Reflect the diagram about the yaxs. See Figure 12.89. 30. (a) See Figure 12.90. y \ b i 1 1 y b i 1 Figure 12.90 Figure 12.91 (b) See Figue 12.91. 31. (a) (i) y (ii) Y x (b) The function f(x, y) = g(y  x) is constant on lies y  x = k. Thus a1lines paralel to y = x are level cures of f. 32. Since f(x,y) = x2  y2 = (x  y)(x + y) = 0 gives x  y = 0 or x + y = 0, the contours f(x,y) = 0 are the lies y = x or y = x. In the regions between them, f(x, y) )0 0 or f(x, y) 0( 0 as shown in Figure 12.92. The surace z = f(x, y) is above the xyplane where f )0 0 (that is on the shaded regions contag the xaxs) and is below the xyplane where f 0( O. Ths means that a person could sit on the surace facing along the positive or negative xaxs, and with hislher legs hangig down the sides below the yaxs. Thus, the graph of the function is saddleshaped at the origi. ...;r. &quot; 12.4 SOLUTIONS 88 y 1 =0 x g=O y=x 1=0 y= x x y=x y= x Figure 12.92 Figure 12.93 33. We need thee lines with g(x, y) = 0, so that the xyplane is divided into six regions. For example g(x, y) = y(x  y)(x + y) has the contour map in Figue 12.93. (Many other answers to ths question are possible.) Solutions for Section 12.4 Exercises 1. (a) Yes. (b) The coeffcient of m is 15 dollars per month. It represents the monthy charge to use this service. The coeffcient of t is 0.05 dollars per miute. Each miute the customer is online costs 5 cents. (c) The intercept represents the base charge. It costs $35 just to get hooked up to ths servce. (d) We have f(3, 800) = 120. A customer who uses ths servce for thee months and is online for a tota of 800 miutes is charged $120. 2. (a) Since z is a liear function of x and y with slope 2 in the xdiection, and slope 3 in the ydiection, we have: z = 2x + 3y + c We can write an equation for changes in z in terms of changes in x and y: L\z = (2(x + L\x) + 3(y + L\y) + c)  (2x + 3y + c) = 2L\x + 3L\y Since L\x = 0.5 and L\y = 0.2, we have L\z = 2(0.5) + 3( 0.2) = 0.4 So a 0.5 change in x and a 0.2 change in y produces a 0.4 change in z. (b) As we know that z = 2 when x = 5 and y = 7, the value of z when x = 4.9 and y = 7.2 wil be z = 2 + L\z = 2 + 2L\x + 3L\y where L\z is the change in z when x changes from 4.9 to 5 and y changes from 7.2 to 7. We have L\x = 4.9  5 =0.1 and L\y = 7.2  7 = 0.2. Therefore, when x = 4.9 and y = 7.2, we have z =2 + 2 . (0.1) + 5 . 0.2 = 2.4 884 Chapter Twelve /SOLUTIONS 3. (a) Substitutig in the values for the slopes, we see that the formula for the plane is z = c + 5x  3y for some value of c. Substituting the point (4, 3,  2) gives c = 13. The formula for the plane is z = 13 + 5x  3y....
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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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