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?

Answer 1

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

Answer 2

Whats the answer