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.
When faced with problems like these, the first step you should always take is to label your variables and make sure you understand the constraints of the problem. Here, the problem tells us that the area of the enclosure will be 4800 yards squared, 280 yards of fencing will be used to enclose the enclosure, the width is not longer than the length, and most importantly that this is a rectangular enclosure. Letting x be the width and y be the length, we'll try to break down what it means in terms of x and y:
The area is at least 4800 yards:
Well, we know it's a rectangle and a rectangle's area is the product of its sides, so in terms of x and y we have:
[tex] xy \geq 4800 [/tex]
280 yards of fencing is to be used:
Again, we can use the fact that it's a rectangle to evaluate this. This constraint is basically telling us that our perimeter is 280 yards, and the perimeter of a rectangle is just the sum of it's sides. That gives us:
[tex] 2(x+y)=280 [/tex]
Lastly we're given that the width "cannot exceed the length". Since x is our width and y is our length, we are basically given:
[tex] x \leq y [/tex]
So, combining this information we have:
[tex] xy \geq 4800\\2(x+y)=280\\x\leq y [/tex]
All that's left to do is to solve the first inequality. Remember, we are looking for x because the problem basically asks us about the range of the width, so that means we can use the second equation to obtain:
[tex] 2(x+y)=280 \implies \\x+y=140 \implies\\y=140-x [/tex]
Since we've solved for y, we can replace the y in the inequality with 140-x to get:
[tex] x(140-x) \geq 4800 [/tex]
I'm going to bring everything to one side and make sure the leading coefficient of the quadratic is positive one, so:
[tex] 140x-x^2 \geq 4800 \implies\\0\geq x^2-140x+4800 [/tex]
So, we're faced with a problem. Should we factor this or do we use the quadratic formula? Luckily for us, we can solve this fairly easy by finding the discriminant of the quadratic. The discriminant is found with the expression: [tex] b^2-4ac [/tex] of a quadratic in standard form. In this case our discriminant would be:
[tex] 140^2-4(4800)=400 [/tex]
So, since the discriminant is a perfect square, our quadratic is factorable. Now, what are the factors? Recall that 6*8=48, and 6+8=14. That tells us that 60*80=4800 and 60+80=140. Since the numbers add to -140 and multiply to positive 4800, they are both negative. So, splitting the -140x up into -60x and -80x gives us:
[tex] 0 \geq x^2-140x+4800 \implies\\0 \geq x^2-80x-60x+4800 \implies\\0 \geq x(x-80)-60(x-80) \implies \\0 \geq (x-60)(x-80) \impliesx\leq 80 \text{ and } x\geq 60 [/tex]
Because of this, we know that x is between 60 and 80.
But, this is where the fact about [tex] x\leq y [/tex], comes in. We are ensured that any x value we pick between 60 and 80 that is paired with a y value that adds to 140 will multiply to a value greater than 4800, so we could always guess and check, but that is unfulfilling. Recall that:
[tex] y=140-x\\x \leq y [/tex]
That means:
[tex] x \leq 140-x \implies\\2x \leq 140 \implies \\x\leq 70 [/tex]
So, with that we know that x lies on the closed interval [60,70].
