To receive full credit for a correct answer please, show your work and processes in solving each question. Solve the system of equations for each variable. 1x + 2y - z = 5 2x - y + 3z = 0 2y + z = 1 2- Two adjoining garden plots have a total area of 256 square feet are enclosed with 80 feet of fencing. One plot is square but the other is a rectangular plot with the length 3 times as large as the width. Determine the length and width of each garden plot. 3- A travel agent offers two package vacation plans. The first plan costs $360 and includes 3 days at a hotel and a rental car for 2 days. The second plan costs $500 and includes 4 days at a hotel and a rental car for 3 days. The daily charge for the hotel is the same under each plan and the daily charge for the car is also the same for each plan. Find the cost per day for the hotel and for the car. 4- A company is planning to produce and sell a new line of TVs. The fixed cost for the company is $400,440 and each TV costs $845 to produce. If they project a sales price of $1200, how many of TVs will need to be sold to break even?? If they decide to reduce the price to $1199.99, how many additional TVs will they need to sell to break even?? 5- A calculator company produces a scientific calculator and a graphing calculator. Long-term projections indicate an expected demand of at least 100 scientific and 80 graphing calculators each day. Because of limitations on production capacity, no more than 200 scientific and 170 graphing calculators can be made daily. To satisfy a shipping contract, a total of at least 200 calculators much be shipped each day. If each scientific calculator sold results in a $2 loss, but each graphing calculator produces a $5 profit, how many of each type should be made daily to maximize net profits? a) b) c) d) e) Define the objective equation. Define the constraint equations. Graph the associated constraint equations. Determine the vertices of each of the graphed equations. Determine how many of each type of calculator should be made daily to maximize the net profits and the amount of net profit. 1) 2) here, drew x by x square adjoined to a w by 3w rectangle, and showed the rectangle longer than the square. There is an unadjoined length on the rectangle of 3w-x. I might later create and upload a picture of this on an image host site. Total Area, w^2 + w.3w=256 4w^2 = 256 Fencing is perimeter , 2x+x+(3w-x+3w+2w)=80 Simplify to 2x+8w =80 x+4w=40 3) Let the cost per day for the room be $x and that of the car be $y. As per the first plan, 3x+2y = 360 As per the second plan, 4x+3y = 500 So the system of equations are 3x+2y = 360 -------------(1) 4x+3y = 500 -------------(2) This is going to be solved by elimination. (1) times 3 gives, 9x+6y = 1080 (2) times 2 gives, 8x+6y = 1000 Subtracting the above 2 equations, x = 80 Plugging in this value in (1), 3(80)+2y = 360 240+2y = 360 ==> 2y = 360 - 240 ==> 2y = 120 ==> y = 60 Thus the cost of a room per day is $80 and that for a car is $60 4) case 1:$400,440 + 845x =1200x 1200x-845x=400440 355x=400440 x=$1128 Case 2:$400,440 + 845x = 1199.99x 1199.99x-845x=400440 354.99x=400440 x=$1128.03 5) The question asks for the optimal number of calculators, so my variables will stand for that: x: number of scientific calculators produced y: number of graphing calculators produced Since they can't produce negative numbers of calculators, I have the two constraints, x > 0 and y > 0. But in this case, I can ignore these constraints, because I already have that x > 100 and y > 80. The exercise also gives maximums: x < 200 and y < 170. The minimum shipping requirement gives me x + y > 200; in other words, y > -x + 200. The profit relation will be my optimization equation: P = -2x + 5y. So the entire system is: P = -2x + 5y, subject to: 100 < x < 200 80 < y < 170 y > -x + 200 corner points at (100, 170), (200, 170), (200, 80), (120, 80), and (100, 100), you should obtain the maximum value of P = 650 at (x, y) = (100, 170). That is, the solution is "100 scientific calculators and 170 graphing calculators". 1) x+2y-z=5 2x-y+3z=0 2y+z=1 Solve for z in 2y+z=1 => Z=1-2y Substitute z=1-2y into x+2y-z=5 => x+4y-1=5 Substitute z=1-2y into 2x-y+3z=0 => 2x-7y +3 =0 Solve for x in x+ 4y -1 =5 => x=-4y +6 Substitute x=-4y+6 into 2x-7y+3=0 => -15y+15 =0 => y=1 Substitute y=1 into 2y+z=1 => 2+z=1 => z=-1 Substitute y=1 and z=-1 into x+2y-z=5 => x+2+1=5 => x=2 X=2;y=1;z=-1 2) here, drew x by x square adjoined to a w by 3w rectangle, and showed the rectangle longer than the square. There is an unadjoined length on the rectangle of 3w-x. I might later create and upload a picture of this on an image host site. Total Area, w^2 + w.3w=256 4w^2 = 256 Fencing is perimeter , 2x+x+(3w-x+3w+2w)=80 Simplify to 2x+8w =80 x+4w=40 3) Let the cost per day for the room be $x and that of the car be $y. As per the first plan, 3x+2y = 360 As per the second plan, 4x+3y = 500 So the system of equations are 3x+2y = 360 -------------(1) 4x+3y = 500 -------------(2) This is going to be solved by elimination. (1) times 3 gives, 9x+6y = 1080 (2) times 2 gives, 8x+6y = 1000 Subtracting the above 2 equations, x = 80 Plugging in this value in (1), 3(80)+2y = 360 240+2y = 360 ==> 2y = 360 - 240 ==> 2y = 120 ==> y = 60 Thus the cost of a room per day is $80 and that for a car is $60 4) case 1:$400,440 + 845x =1200x 1200x-845x=400440 355x=400440 x=$1128 Case 2:$400,440 + 845x = 1199.99x 1199.99x-845x=400440 354.99x=400440 x=$1128.03 5) The question asks for the optimal number of calculators, so my variables will stand for that: x: number of scientific calculators produced y: number of graphing calculators produced Since they can't produce negative numbers of calculators, I have the two constraints, x > 0 and y > 0. But in this case, I can ignore these constraints, because I already have that x > 100 and y > 80. The exercise also gives maximums: x < 200 and y < 170. The minimum shipping requirement gives me x + y > 200; in other words, y > -x + 200. The profit relation will be my optimization equation: P = -2x + 5y. So the entire system is: P = -2x + 5y, subject to: 100 < x < 200 80 < y < 170 y > -x + 200 corner points at (100, 170), (200, 170), (200, 80), (120, 80), and (100, 100), you should obtain the maximum value of P = 650 at (x, y) = (100, 170). That is, the solution is "100 scientific calculators and 170 graphing calculators". 1) 2) here, drew x by x square adjoined to a w by 3w rectangle, and showed the rectangle longer than the square. There is an unadjoined length on the rectangle of 3w-x. I might later create and upload a picture of this on an image host site. Total Area, w^2 + w.3w=256 4w^2 = 256 Fencing is perimeter , 2x+x+(3w-x+3w+2w)=80 Simplify to 2x+8w =80 x+4w=40 3) Let the cost per day for the room be $x and that of the car be $y. As per the first plan, 3x+2y = 360 As per the second plan, 4x+3y = 500 So the system of equations are 3x+2y = 360 -------------(1) 4x+3y = 500 -------------(2) This is going to be solved by elimination. (1) times 3 gives, 9x+6y = 1080 (2) times 2 gives, 8x+6y = 1000 Subtracting the above 2 equations, x = 80 Plugging in this value in (1), 3(80)+2y = 360 240+2y = 360 ==> 2y = 360 - 240 ==> 2y = 120 ==> y = 60 Thus the cost of a room per day is $80 and that for a car is $60 4) case 1:$400,440 + 845x =1200x 1200x-845x=400440 355x=400440 x=$1128 Case 2:$400,440 + 845x = 1199.99x 1199.99x-845x=400440 354.99x=400440 x=$1128.03 5) The question asks for the optimal number of calculators, so my variables will stand for that: x: number of scientific calculators produced y: number of graphing calculators produced Since they can't produce negative numbers of calculators, I have the two constraints, x > 0 and y > 0. But in this case, I can ignore these constraints, because I already have that x > 100 and y > 80. The exercise also gives maximums: x < 200 and y < 170. The minimum shipping requirement gives me x + y > 200; in other words, y > -x + 200. The profit relation will be my optimization equation: P = -2x + 5y. So the entire system is: P = -2x + 5y, subject to: 100 < x < 200 80 < y < 170 y > -x + 200 corner points at (100, 170), (200, 170), (200, 80), (120, 80), and (100, 100), you should obtain the maximum value of P = 650 at (x, y) = (100, 170). That is, the solution is "100 scientific calculators and 170 graphing calculators". 1) x+2y-z=5 2x-y+3z=0 2y+z=1 Solve for z in 2y+z=1 => Z=1-2y Substitute z=1-2y into x+2y-z=5 => x+4y-1=5 Substitute z=1-2y into 2x-y+3z=0 => 2x-7y +3 =0 Solve for x in x+ 4y -1 =5 => x=-4y +6 Substitute x=-4y+6 into 2x-7y+3=0 => -15y+15 =0 => y=1 Substitute y=1 into 2y+z=1 => 2+z=1 => z=-1 Substitute y=1 and z=-1 into x+2y-z=5 => x+2+1=5 => x=2 X=2;y=1;z=-1 2) here, drew x by x square adjoined to a w by 3w rectangle, and showed the rectangle longer than the square. There is an unadjoined length on the rectangle of 3w-x. I might later create and upload a picture of this on an image host site. Total Area, w^2 + w.3w=256 4w^2 = 256 Fencing is perimeter , 2x+x+(3w-x+3w+2w)=80 Simplify to 2x+8w =80 x+4w=40 3) Let the cost per day for the room be $x and that of the car be $y. As per the first plan, 3x+2y = 360 As per the second plan, 4x+3y = 500 So the system of equations are 3x+2y = 360 -------------(1) 4x+3y = 500 -------------(2) This is going to be solved by elimination. (1) times 3 gives, 9x+6y = 1080 (2) times 2 gives, 8x+6y = 1000 Subtracting the above 2 equations, x = 80 Plugging in this value in (1), 3(80)+2y = 360 240+2y = 360 ==> 2y = 360 - 240 ==> 2y = 120 ==> y = 60 Thus the cost of a room per day is $80 and that for a car is $60 4) case 1:$400,440 + 845x =1200x 1200x-845x=400440 355x=400440 x=$1128 Case 2:$400,440 + 845x = 1199.99x 1199.99x-845x=400440 354.99x=400440 x=$1128.03 5) The question asks for the optimal number of calculators, so my variables will stand for that: x: number of scientific calculators produced y: number of graphing calculators produced Since they can't produce negative numbers of calculators, I have the two constraints, x > 0 and y > 0. But in this case, I can ignore these constraints, because I already have that x > 100 and y > 80. The exercise also gives maximums: x < 200 and y < 170. The minimum shipping requirement gives me x + y > 200; in other words, y > -x + 200. The profit relation will be my optimization equation: P = -2x + 5y. So the entire system is: P = -2x + 5y, subject to: 100 < x < 200 80 < y < 170 y > -x + 200 corner points at (100, 170), (200, 170), (200, 80), (120, 80), and (100, 100), you should obtain the maximum value of P = 650 at (x, y) = (100, 170). That is, the solution is "100 scientific calculators and 170 graphing calculators"