SOLUTION: An existing landfill is built on a rectangular parcel of property. The width of the landfill is 100m. If the landfill permit is changed is changed to increase the length of the lan

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Question 883589: An existing landfill is built on a rectangular parcel of property. The width of the landfill is 100m. If the landfill permit is changed is changed to increase the length of the landfill by 50m, its area will be increased by 5000 square meters (m^2) and its useful life span will be extended by 20 percent. What is the current length of the landfill.
Since this is an area problem, I'm guessing the 20% life span in inconsequential, so I'm ignoring it.
I use A=WL and get as far as this:
(100)(L + 50) = A + 5000
This really doesn't seem to take me anywhere. Any help is much appreciated.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
THE EFFICIENT SOLUTION (with my usual chatty comments/explanations):
Comment: If they add 50 meter to the length,
and the width is 100 meters,
they are adding a 20 meter by 100 meter rectangle to one end of the rectangular landfill.
The added rectangle has a surface area of
%28100m%29%2850m%29=5000m%5E2 , of course.
You did not need to be told what the are increase would be,
but since they did, you did not have to multiply.
Anyway, you do not need that number.

Solution (with reasoning):
If you change the length of a rectangle by any factor,
but leave the width unchanged, the area changes by that factor.
The enlargement adds %2220+%25%22=20%2F100=0.20=1%2F5 to the useful life span of the landfill
because the enlarged area is increased by that factor,
and the area increased by that factor
because the length is increased by that factor.
The 50m to be added are 1%2F5 of the current/original length,
The current length is 5%2A50m=highlight%28250m%29 .

More comment: The 20% and 50m information is all you needed.
The width and the area added are irrelevant.
Unfortunately, your teacher may not know that, and may not be able to understand that right away, so maybe the expected solution is the inefficient one.

THE SOLUTION THAT MAY BE EXPECTED:
L= length of the landfill, in meters.
Since the landfill is rectangular and the width is 100 meters.
100L= current surface area of the landfill, in square meters.

---(HERE YOU COULD INSERT THE UNNECESSARY PART IF THEY INSIST)---

If that increase in area is 20% of the current area,
20%2F100=5000%2F100L-->20%28100L%29=5000%2A100-->2000L=500000-->L=5000000%2F20000=highlight%28250%29

THE PART THAT FOLLOWS IS TRUE, BUT REALLY UNNECESSARY:
L%2B50= planned new length of the landfill, in meters.
Since the enlarged landfill would still be rectangular, and its width would still be 100 meters,
100%28L%2B50%29=100L%2B100%2A50=100L%2B5000= surface area of the enlarged landfill, in square meters.
Compared to the current area, 100Lm%5E2 , the enlarged landfill area, 100L%2B5000m%5E2 is indeed 5000m%5E2 larger).