Week II Assignment

22 November 2012

This week assignment required to solve problem 68 on page 539 of Elementary and Intermediate Algebra. There are three part to this assignment and the first part is as follows; The accompanying graph shows all of the possibilities for the number of refrigerators and the number of TVs that will fit into an 18-wheeler. | |

a)| Write an inequality to describe this region.|

p = y1-y2 / X1-x2 = 330 – 0 / 0-110 = -3/1 the slope is -3/1 or -3 Y – y1 = p(x – x1)

Y – 330 = - 3 / 1(x-0)

Y = - 3x/1 + 330

-3x/1 +330 = y expression switch by place the y on the right hand side -3x/-3 = y/-3 – 330/ -3 divide each equation by -3 and cancel out like terms -3y = 1x + 110

-3y + 1x < 110

b)| Will the truck hold 71 refrigerators and 118 TVs?|

-3 (71) + 1 (118) < 110

-213 + 118 < 110

-95 < 110 no the truck will not hold 71 refrigerators and 118 TVs. c)| Will the truck hold 51 refrigerators and 176 TVs?|

-3 (51) + 1 (176) < 110

-153 +176 < 110

23 < 110 yes the truck will hold at least 51 refrigerators and 176 TVs

The Burbank Buy More store is going to make an order which will include at most 60 refrigerators. What is the maximum number of TVs which could also be delivered on the same 18-wheeler? Describe the restrictions this would add to the existing graph. Solving for y

1(60) + -3y < 110

-3y < -60 + 110 add 110 to -60 to get 50

-3y < 50 divide both terms by -3

-3y/-3 > 50/-3 signs flip

Y > -50/3 or y = 16 there will be no added restriction because the maximum numbers of TVs The next day, the Burbank Buy More decides they will have a television sale so they change their order to include at least 200 TVs. What is the maximum number of refrigerators which could also be delivered in the same truck? Describe the restrictions this would add to the original graph. 1x + -3 (200) < 110

X < 600 + 100

X = < 710

If 200 TVs are ship then...

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