Question

MEDAL AWARD. At most 500 ft of fencing is available for a steer pasture. There is only space for the pasture to be at most 100 ft wide. Write a system of two linear inequalities that describes this situation and write two possible solutions.

Answer 1

Let's assume we have a rectangle

Let x be the width and y be the length

Answer 2

It states that "At most 500 ft of fencing is available for a steer pasture"

so the perimeter P is at most 500 ft, which means \[\Large P \le 500\]

Answer 3

The perimeter of the rectangle above is P = 2(x+y), so...

\[\Large P \le 500\]

\[\Large 2(x+y) \le 500\]

\[\Large x+y \le \frac{500}{2}\]

\[\Large x+y \le 250\]

Answer 4

"There is only space for the pasture to be at most 100 ft wide" means that \[\Large x \le 100\]

Answer 5

And finally, the values of x and y cannot be negative (since they are the width and length) telling us that

\[\Large x \ge 0\]
\[\Large y \ge 0\]

Answer 6

Put this all together and you get this system of inequalities

\[\Large x+y \le 250 \\ \Large x \le 100 \\ \Large x \ge 0 \\ \Large y \ge 0\]

Answer 7

I'm not following this.. There is only supposed to be 2 linear inequalities.. And when you solved p is less than or equal to 500, wouldn't you come out with the equation, y= -x+250?

Answer 8

ok, so you can ignore the last two inequalities if you want and just focus on quadrant 1

Answer 9

so you'd have the system

\[\Large x+y \le 250 \\ \Large x \le 100\]

Answer 10

And yes, you can solve for y to get

\[\Large y \le -x+250\]

Answer 11

the equation form of that is y = -x + 250

So you would graph the equation to form the boundary line