A rectangular enclosure must have an are of at least 4800yd^2. If 280yd of fencing is used, and the width can not exceed the length, within what limits must the width of the enclosure lie?
let the width be x, and the length y
first restriction: y ≥ x
2nd restriction: 2x + 2y ≤ 280
x+y ≤ 140
y ≤ 140 - x
3rd restriction: xy ≥ 4800
y ≥ 4800/x
On an x-y grid using only the first quadrant :
draw the line y = x, and shade in the part above the line
draw the line y = 140 - x , and shade in the part below the line
sketch the curve y = 4800/x and shade in the part above the curve
http://www.wolframalpha.com/input/?i=plot++y+%3D+4800%2Fx+,+y+%3D+140-x,+y+%3D+x+,+from+0+to+140
(notice the scaling in not 1:1)
the area of interest seems to be shaded part bounded by the intersection of
y = 4800/x and y = 140-x, y = 4800/x and y = x, and y = x and y = 140-x
here is a "close-up" of your region
http://www.wolframalpha.com/input/?i=plot++y+%3D+4800%2Fx+,+y+%3D+140-x,+y+%3D+x+,+from+55+to+80
Here are the solutions for those intersections:
http://www.wolframalpha.com/input/?i=solve++y+%3D+4800%2Fx+,+y+%3D+140-x
http://www.wolframalpha.com/input/?i=solve++y+%3D+4800%2Fx+,+y+%3D+x
http://www.wolframalpha.com/input/?i=solve++y+%3D+140-x+,+y+%3D+x
draw your conclusion from this analysis
check by picking a value of x outside your stated domain, and a value of x within your domain
Whats the answer
Let's assume that the length of the enclosure is represented by L and the width is represented by W.
Given that the perimeter of the enclosure is 280yd, we can set up an equation as follows:
2L + 2W = 280
Since the width cannot exceed the length, we know that W ≤ L.
To find the limits within which the width must lie, we'll use the fact that the area of a rectangle is given by the formula:
Area = Length * Width
Given that the area must be at least 4800yd^2, we can set up an inequality as follows:
L * W ≥ 4800
Now, let's solve these equations together step-by-step.
Step 1: Solve the perimeter equation for L:
2L + 2W = 280
Simplify the equation by dividing both sides by 2:
L + W = 140
Solve for L:
L = 140 - W
Step 2: Substitute L into the area inequality equation:
(140 - W) * W ≥ 4800
Multiply out the left side:
140W - W^2 ≥ 4800
Rearrange the equation:
W^2 - 140W + 4800 ≤ 0
Step 3: Find the roots of the quadratic equation:
To find the limits within which W must lie, we need to find the values of W that make the inequality true. Set the quadratic equation equal to zero:
W^2 - 140W + 4800 = 0
Factoring the equation:
(W - 80)(W - 60) = 0
Setting each factor equal to zero:
W - 80 = 0 or W - 60 = 0
Solving for W:
W = 80 or W = 60
Step 4: Determine the limits for the width:
Since W cannot exceed L, we need to consider the case where W = L and L = W.
For W = 80, L = 80, and for W = 60, L = 60.
Therefore, the limits for the width of the enclosure lie between 60 yards and 80 yards.
To solve this problem, we need to set up an equation based on the given constraints and then solve for the range of possible values for the width of the enclosure.
Let's assume that the length of the enclosure is L and the width is W.
Based on the first constraint, the area of the enclosure must be at least 4800 yd². We can write this inequality as:
L * W ≥ 4800
Now, let's consider the second constraint. The total amount of fencing used is 280 yd. Since the enclosure is rectangular, the amount of fencing required would be twice the length plus twice the width:
2L + 2W = 280
Since the width cannot exceed the length, we can write another inequality:
W ≤ L
Now, we have a system of inequalities:
L * W ≥ 4800
2L + 2W = 280
W ≤ L
To find the range of possible values for the width of the enclosure, we need to solve this system of inequalities.
First, let's solve the second equation for L:
2L + 2W = 280
2L = 280 - 2W
L = 140 - W
Next, substitute this expression for L in the first and third inequality:
W * (140 - W) ≥ 4800
W ≤ 140 - W
Now, we can simplify these inequalities:
-W² + 140W ≥ 4800
2W ≤ 140
Rearranging the first inequality:
W² - 140W + 4800 ≤ 0
To find the range of possible values for W, we need to find the values for which this quadratic inequality is true. We can solve this by factoring or using the quadratic formula.
By factoring, we get:
(W - 60)(W - 80) ≤ 0
This inequality is true when either:
W - 60 ≤ 0 and W - 80 ≥ 0
or
W - 60 ≥ 0 and W - 80 ≤ 0
From the first case, we find:
W ≤ 60 and W ≥ 80
From the second case, we find:
W ≥ 60 and W ≤ 80
Combining both cases, we find that the width, W, must lie within the limits:
60 ≤ W ≤ 80
Therefore, the width of the enclosure must be between 60 yards and 80 yards, inclusive.