Question 1207290: A veterinarian wishes to use 168 feet of chain-link fencing to enclose a rectangular region and subdivide the region into two smaller rectangular regions, as shown in the following figure. If the total enclosed area is 1026 square feet, find the width w and length l of the enclosed region.
w = ft (smaller width) by l = ft
w = ft (larger width) by l = ft
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! as far as i can tell, the enclosed region has 2 lengths and 3 widths.
2 lengths and 2 widths are on the outside of the enclosed region.
one width splits the enclosed region into two smaller regions that are equal in area.
i get the following two possibilities.
the length is 27 and the width is 38.
the length is 18 and the width if 57.
length * width = 1026 in both cases.
2 lengths and 3 widths = 168 in both cases.
my equations were:
length * width = 1026
2 * length + 3 * width = 168.
i solved for length * width from length * width = 1026.
this led to length = width / 1026.
i substituted for length in 2 * length + 3 * width = 168 to get:
2 * 1026 / width + 3 * width = 168.
i multiplied both sides of taht equation by width to get:
2 * 1026 + 3 * width squared = 168 * width.
i subtracted 168 * width from both sides of the equation to get:
2 * 1026 + 3 * width squared - 168 * width = 0
i simplified and order in descending order of degree to get:
3 * width squared - 168 * width + 2052 = 0
i factored this quadratic equation to get width = 38 or 18.
when width was 38, length was 27.
when width was 18, length was 57.
length * width = 1026 in both equations.
2 * length + 3 * width = 168 in both equtions.
those appeared to be two feasible solutions to the problem as i understood it.
my diagram looked like this:
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