The length and width of a book cover are 22.2 centimeters and 12 centimeters respectively. The actual length (and width) can be 0.3 unit less than the measured length (and width) or 0.3 unit greater than the measured length (and width). a. Find the minimum and maximum possible lengths and widths of the book cover. b. Calculate the minimum and maximum possible areas of the book cover.

Part (a)

The length is supposed to be 22.2 cm, but it could be 0.3 cm less

So 22.2 - 0.3 = 21.9 cm is the smallest value for the length. This is the lower bound of the length.

The upper bound is 22.2 + 0.3 = 22.5 cm as it is the largest the length can get.

Use this for the width as well

The width is supposed to be 12 cm, but it could be as small as 12-0.3 = 11.7 cm and as large as 12+0.3 = 12.3 cm

Part (b)

Use the smallest length and width to get the smallest possible area

smallest area = (smallest width)*(smallest length) = 11.7*21.9 = 256.23

Repeat the same idea but for the largest length and width to get the largest possible area

largest area = (largest width)*(largest length) = 12.3*22.5 = 276.75

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3