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5, Note: These situations are shown only for the first quadrant. (a) x24 The graph of x =4 represents a vertical line with x-intercept (4,0). Since any value of x to the right of this line is greater than 4, we shade the portion to the right of this vertical line. Ss (4,0) (0,0) (b) y29 The graph of y=9 represents a horizontal line with y-intercept (0,9). Since any value of y above this line is greater than 9, we shade the portion above this horizontal line. 7 ners (0,0) 2 (c) 3x+2y<18 _ The y-intercept of 3x+2y =18 can be found by substituting x= 0. 3(0)+2y =18 > 0+2y =18> 2y=18 > y=FH9 The y-intercept is (0,9). The x-intercept of 3x+2y =18 can be found by substituting y = 0. 3x +2(0)=18 > 3x40=18 => 3x=18 >x=F=6 The x-intercept is (6,0). We draw a line connecting these points. Testing the point (0,0) ,we have the statement 3(0)+2(0)<18 or OS18. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line. Be Sx t 2y=16 8 12x +10y < 640 (beef) 2 4x+ 3y $480 (pork) 0.0 2 4 "3 B 1 12x (2.0) Seip (d) 7x+2y<42 The y-intercept of 7x+2y=42 can be found by substituting x =0, 7(0)+2y = 42 = 0+2y =42 => 2p=42 > y=%=71 The y-intercept is (0,21). : The x-intercept of 7x+2y=42 can be found by substituting y= 0. 7x+2(0)=42 > 7x+0=42> 7x=425 x= 256 The x-intercept is (6,0). We draw a line connecting these points. Testing the point (0,0), we have the statement 7(0)+2(0)< 42 or 0<42, This is a true statement, thus we shade the half-plane containing our test point, the down side of the line. ¥ (0, 21) Txt ay=42 (0, 0} (6,0) “x 13. x20; y24;x+y 520 The constraints of x20andy24 indicate that we are restricted to the upper right quadrant, above the horizontal line y= 4. The point of intersection between y= 4 and x+ y= 20 can be found by substituting y=4 into eye 20: x+4=20>x=16 Thus, the point of intersection is (16,4). The y-intercept of x+y =20 can be found by substituting x = 0. 0+ y=20>5 y=20 The y-intercept is (0,20). The x-intercept of x+y = 20 can be found by substituting y =0. x+0= 20> x=20 The x-intercept is (20,0). We draw a line connecting these points. Testing the point (0,0},we have the statement 0+0<20 or 0<20. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line, which is contained in the upper right quadrant above the horizontal line y = 4, ¥ (0,20) |<—x=0 20 16 12 (0,4) 8 4 0) 4 3 12 %f2O x (20, 0) 19. Note: These situations are shown only for the first quadrant. (a) Sx+4y=22 and 5x+10y=40 The y-intercept of 5x+4y = 22 can be found by substituting x = 0. 5(0)+4y = 22 O+4y=22 4y = 22 yoBe=5.5 The y-intercept is (0,5.5). The x-intercept of S5x+4y=22 can be found by substituting y = 0. 5x+4(0) =22 5x+0=22 5x=22 x=Bea4 The x-intercept is (4.4,0). The y-intercept of Sx+10y = 40 can be found by substituting x = 0. 5(0) +10 =40 0+10y =40 yepa4 The y-intercept is (0,4). The x-intercept of 5x+10y = 40 can be found by substituting y =0. 5x+10(0)= 40 5x+0=40 5x=40 xaB=B The x-intercept is (8,0). To find the point of intersection, we can multiply both sides of 5x +4 y= 22 by —1, and adding the result to 5x+10y = 40. —3x—4y =~22 $x+10y = 40 6y=1 o> y=E=3 Substitute y=3 into 5x+10y=40 and solve for x. 5x+10(3) = 40 > 5x4+30=40= $x=105 x=2=2 The point of intersection is therefore (2,3). (440) 60) (b) x+y=7 and 3x+4yp= 24 The y-intercept of x+y =7 can be found by substituting x = 0. O+y=7 y=T The y-intercept is (0,7). The x-intercept of x+y =7 can be found by substituting y =0. x+0=7 x=7 The x-intercept is (7,0). The y-intercept of 3x+4y=24 can be found by substituting x = 0. , 3(0}+4y = 24 O+4y=24 i The p-imtercept is (0, 6). The x-intercept of 3x+4y =24 can be found by substituting y = 0. 3x+4(0) = 24 3x+0= 24 3x=24 x=H=8 The x-intercept is (8,0). To find the point of intersection, we can multiply both sides of x+ y=7 by —3, and adding the result to 3x+4y = 24 -—3x-—3y =-21 3x+4y = 24 y=3 Substitute y=3 into 3x+4y=24 and solve for x. 3x+4(3) = 24 = 3x+12=24=593x=l2 > x=B=4 The point of intersection is therefore (4,3). (0.0) 1 2 32 4 5 6/7 BX (7,0) 22, x20; p>2;5xn+y Si4;x+2ys10 The constraints of x2Qandy>2 indicate that we are restricted to the upper nght quadrant, above the horizontal line py = 2. The y-intercept of 5x+ y=14 can be found by substituting x =0. 5(0)+yal4o0+ypal>y=14 The y-intercept is (0,14). The x-intercept of 5x+ y =14 can be found by substituting y= 0. 5x+0=14> $x=14> x=4=2.8 The x-intercept is (2.8,0). We draw a line connecting these points. Testing the point (0,0),we have the statement 5(0)+0514 or 0<14. This is a true statement. The y-intercept of x+2y =10 can be found by substituting x =0. O+2y=10> 2y =10> ya =s The y-intercept is (0,5). The x-intercept of x+2y=10 can be found by substituting y = 0. x+2(0)=10>x+0=10>5 x=10 The x-intercept is (10,0). We draw a line connecting these points. Testing the point (0,0),we have the statement O+ 2{0} S10 or O<10. This is a true statement. Thus, we shade the part of the plane in the upper right quadrant which is on the down side of both the lines 5x+y=14 and x+2y=10, which is also above the horizontal line yp =2. Two of the comer points, (0,2) and (0,5), lie on the coordinate axes. The third comer point is the point of intersection between the lines 5x+p=14 and y=2. We can find this by _ Substituting y=2 into 5x+y=14 to solve for y. We have 5x+2=14> 5x=12>x=2=24, The point of intersection is therefore (2.4,2). The fourth comer point is the point of intersection between the lines 5x+y=14 and x+2y=10. We can find this by multiplying both sides of 5x+y=14 by —2, and adding the result to x+2y=10. —10x-2y=-28 x+2y=10 ~9x =-18x=—8=2 Substitute x=2 into x+2y=10 and solve fory. We have 2+2y=10> 2p=8> y=2=4. The point of intersection is therefore (2,4). (0.0) 1 af 45678901 * (2.8, 0) 28. (a) x20; y20;7x+4y<13 The constraints of x 20 and y 20 indicate that we are restricted to the upper right quadrant created by the x-axis and y-axis, The y-intercept of 7x+4y =13 can be found by substituting x=0. 7(O)+4y=13-0+4y=13 > 4y=13 > y=8 The y-intercept is (0,2). For graphing purposes, treat 2 as 31. The x-intercept of 7x+4y=13 can be found by substituting y = 0. Txt 4(0)=13> 7x4+0=13 > 7x=133x=8 The x-intercept is (2,0). For graphing purposes, treat 2 as 14. We draw a line connecting these points. Testing the point (0,0) , We have the statement 7(0)+4(0)<13 or 0513. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line, which is contained in the upper right quadrant. ; (07,0) 2 4 6% (b) According to the optimal production policy, we are looking for the maximum of P=2\ix+lly, given our constraints. Comer Point Value of the Profit Formula: 21x+liy (0,0) 21(0) + 10} = 0 + O= 0 (0,2) 21(0) + Hf¥) = 0 + 8 = 353 i a oO + oa (7.0) 214) + _11(6) The maximum value occurs at the comer point (8, 0). = 39* 33. (a) Letx be the number of routine visits and y be the number of comprehensive visits. Doctor Time (1800 min) Minimums Profit Routine, x visits 5 0 $30 Comprehensive, y visits 25 0 $50 (b) Profit formula: P =$30x+$50y Constraints: x20 and y 20 (minimums); 5x +25y < 1800 (time) {c)} Feasible region: +— x=0 5x + 2By = 1800 (2.72) 20 (0,0) 40 (360.0) *~ 33. (c) contimmed Corner points: The corner points are intercepts on the axes. These are (0,0), (0,72), and (360, 0). {d) We wish to maximize $30x+$50y. Comer Point Value of the Profit Formula: $30x+$50y (0,0) $30(0) + $50(0) = $0 + $0 = $0 (0,72) $30(0) + $50{72) = $0 + $3600 = $3600 (360,0) $30(360) + $50(0) = $10,800 + $0 = $10,800* Optimal production policy: Schedule 360 routine visits and no comprehensive visits. With non-zero minimums, the constraints are as follows. x2 20 and y 230 (minimums);5x+25y < 1800 (time) The feasible region looks like the following. ¥ —— x=20 (0,72) Bx + 25y = 1800 4—~ (20, 68) (210, 30) ef 20 — y=30 (2,0) 40 (20,30) (60,0) * Corner points: One corner point is the point of intersection between x = 20 and y=30, namely (20,30). Another is the point of intersection between x= 20 and 5x+25 y =1800. Substituting x= 20 into 5x+25y = 1800, we have the following, 5(20)+25y =1800 > 1004+25y =1800 > 25y =1700 > y=68 Thus, (20,68) is the second comer point. The third comer point is the point of intersection between y= 30 and 5x+25y = 1800. Substituting y= 30 into $x+25y=1800, we have the following. 5x+25(30) = 1800 => 5x+750 = 18005 $x =1050 > x=210 Thus, (210,30) is the third comer point. We wish to maximize $30x + $50y. Corner Point Value of the Profit Formula: $30x + $50y (20,30) $30(20) + $50(30) = $600 + $1500 = $2100 (20, 68} $30(20) + $50(68} = $600 + $3400 = $4000 (210,30) $30(210) + $50(30) $6300 + $1500 = $7800* Optimal production policy: Schedule 210 routine visits and 30 comprehensive visits. 1 a Supplies o 3 Demands Demands Demands Ee 20 | : Demands 3 L O He. 2 il i <8) S Supplies d Supplies oe a L Demands Demands 
