MHB Find Width of Rectangle with 600yd2 and 140yd Fencing

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To determine the width of a rectangular enclosure with an area of at least 600 yd² and 140 yd of fencing, the width must be less than or equal to the length. The constraints lead to the inequality (70 - w)w ≥ 600, which simplifies to finding valid width values. The width must be less than the square root of 600, indicating it cannot exceed 25 yd. Given the conditions, the only feasible width is 10 yd, resulting in a length of 60 yd, which meets all criteria. Thus, the width must be 10 yd or greater, but not exceeding 25 yd.
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A rectangular enclosure must have an area of at least 600 yd2. If 140 yd of fencing is to be used, and the width cannot exceed the length, within what limits must the width of the enclosure lie?
Select one:
A. 35 ≤ w ≤ 60
B. 10 ≤ w ≤ 35
C. 10 ≤ w ≤ 60
D. 0 ≤ w ≤ 10
 
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length, $L$ and width $(70-L)$

constraints ...

$70-L \le L$

$L(70-L) \ge 600$

see what you can do from here
 
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Since the problem specifically asks about "w", the width, I would, instead, say that $(70- w)w\ge 600$.

Of course that's exactly the same as the inequality skeeter gave for L. There will be two solutions. It is the condition that "the width cannot exceed the length" that allows you to distinguish between them.
 
Width being less than square root of 600 I see that it is less than 25 so I have only one candidate d to check and width 10 so length 60 meets the criteria and hence the answer
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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