GRAPHS, FUR1 2004 VCAA 8 MC

The shaded region shown in the graph above (with boundaries included) is described by

A.	`3x + 4y ≤ 12`
	`x -y ≤ 1`
	`x ≥ 0`
	`y ≥ 0`

B.	`3x + 4y ≤ 12`
	`x -y ≥ 1`
	`x ≥ 0`
	`y ≥ 0`

C.	`3x + 4y ≥ 12`
	`x -y ≥ 1`
	`x ≥ 0`
	`y ≥ 0`

D.	`4x + 3y ≤ 12`
	`x -y ≤ 1`
	`x ≥ 0`
	`y ≥ 0`

E.	`4x + 3y ≤ 12`
	`x -y ≥ 1`
	`x ≥ 0`
	`y ≥ 0`

Show Answers Only

`B`

Show Worked Solution

`text(From the graph)`

`text(Horizontal boundary:)\ y >= 0`

`text(For)\ 1^text(st)\ text(line to the left)`

`text(Line passes)\ (0, text(−1))\ text(and)\ (1, 0)`

`text(Using )`	`(y − y_1)/(x − x_1)`	`= (y_2 − y_1)/(x_2 − x_1)`
	`(y − (text(−1)))/(x − (0))`	`= ((0) − (text(−1)))/((1) − (0))`
	`(y + 1)/x`	`= 1/1`
	`y + 1`	`= x`
	`x − y`	`= 1`

`text(From the graph,)\ x − y >= 1`

`text(For)\ 2^text(nd)\ text(line to the right)`

`text{Line passes (4, 0) and (0, 3)}`

`text(Using )`	`(y − y_1)/(x − x_1)`	`= (y_2 − y_1)/(x_2 − x_1)`
	`(y − (0))/(x − (4))`	`= ((3) − (0))/((0) − (4))`
	`y/(x − 4)`	`= 3/text(−4)`
	`text(−4)y`	`= 3x − 12`
	`text(−3)x − 4y`	`= text(−12)`
	`3x + 4y`	`= 12`

`text(From the graph)`

`3x + 4y <= 12`

`=> B`