Arectangular enclosure must have an area of at least 4800 yd2. if 280 yd of fencing is to be used, and the width cannot exceed the length, within what limits must the width of the enclosure lie?

The problem statement gives rise to three inequalities. If we let x represent the length and y represent the width, we have

... x·y ≥ 4800 . . . . . area constraint

... 2(x + y) ≤ 280 . . . perimeter constraint

... x ≥ y . . . . . . . . . . dimension constraint

These are graphed below.

We can see from the graph that the maximum value of y will be found where x=y and 2(x+y)=280, or

... 4y = 280

... y = 70 . . . . yards

The minimum value of x will be where x=140-x and xy=4800, or

... y(140-y) = 4800

... y² -140y + 4800 = 0

... (y -60)(y -80) = 0

... y = 60 . . . . . . minimum of the solutions to this.

The range the width may have is [60, 70] yards.

part 1 answer = 8

part 2 answer = 72

step-by-step explanation:

part 1 explanation = add all of the numbers, then divide by how many numbers you have. so, 200/25 = 8.

part 2 explanation = not gonna lie, this is hard to explain, so just google how to do it.