SOLUTION: A farmer has a 100foot by 25 foot rectangular field that he wants to reduce to 16% of its original size. How wide of a strip should he cut around the edge of his field to do this?

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: A farmer has a 100foot by 25 foot rectangular field that he wants to reduce to 16% of its original size. How wide of a strip should he cut around the edge of his field to do this?       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 1166543: A farmer has a 100foot by 25 foot rectangular field that he wants to reduce to 16% of its original size. How wide of a strip should he cut around the edge of his field to do this?
Found 3 solutions by htmentor, ikleyn, greenestamps:
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
Let l, w be the length and width of the lawn. Let s = the strip width.
The length of the rectangle remaining to mow is l - 2s, since it's reduced by s on either side
Similarly, the width remaining to mow is w - 2s
The new area is 16% of the original area:
A1 = 0.16*l*w = 0.16*100*25 = 400 = (l-2s)(w-2s)
(100-2s)(25-2s) = 400
2500 - 200s - 50s + 4s^2 - 400 = 0
4s^2 - 250s + 2100 = 0
2s^2 - 125s + 1050 = 0
This can be factored as: (2s-105)(s-10) = 0
The first solution gives a strip wider than the width of the lawn, so we take
the 2nd: s=10
A 10 ft wide strip will leave 16% of the original area
Check: (100-20)(25-20) = 80*5 = 400

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

(100-2x)*(25-2x) = 0.16*100*25    (1)    (the new area equation)


(100-2x)*(25-2x) = 400 = 80*5


the numbers 80 and 5 are two factors / (divisors) of 400 that differ by 80-5 = 75,  same as the factors (100-2x) and (25-2x).


ANSWER.  The new dimensions are 80 and 5 feet.

Solved.

Alternatively, you may solve the quadratic equation (1).

--------------

To see many similar problems on a rectangle and a surrounding strip of a uniform width,
look into the lessons
    - Problems on the area and the dimensions of a rectangle surrounded by a strip
    - Cynthia Besch wants to buy a rug for a room
    - OVERVIEW of lessons on dimensions and the area of rectangles and circles and their elements
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic
"Dimensions and the area of rectangles and circles and their elements".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.




Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Just some comments about the two responses you have received up until now....

The first tutor showed a pure algebraic solution to the problem. It goes something like this....

0.16%28100%2A25%29+=+%28100-2x%29%2825-2x%29
400+=+2500-250x%2B4x%5E2
4x%5E2-250x%2B2100+=+0
2x%5E2-125x%2B1050+=+0
%282x-105%29%28x-10%29+=+0
x+=+52.5 or x+=+10

The first solution to the equation is rejected, because it makes no sense in the original problem. So the answer x=10.

ANSWER: He cut a strip 10 feet wide around the edge to reduce the field to 16% of its original size.

That's a perfectly good formal algebraic solution to the problem. But along the way, there was one very difficult step -- factoring the quadratic into two linear factors. That was an ugly quadratic to factor; doing it takes most students a large amount of time.

So we got a solution by that method -- but it took a lot of work.

The second tutor essentially said "do a bit of THINKING to find the answer". Or, if you want, you can find the answer by solving the equation.

(Oh, NO! She's asking you to THINK!)

Yes, you should know how to solve this problem using formal algebra. However, if you want to reach the answer as quickly as possible, and with as little work as possible, try playing with some numbers. It will give you some good mental exercise; and you will find the mental arithmetic you need to do is MUCH less than the work you had to do to factor that quadratic.

You are cutting a strip x units wide from a 100x25 foot field to end up with a field with an area of 400 square feet.

The other tutor suggested the strategy of looking for two numbers that differ by 100-25=75 and whose product is 400. That's one good strategy that might work well for many students.

Here is another: Try different values for the width of the strip and find which one gives you a field with an area of 400 square feet.

5 feet wide: (100-10)(25-10) = 90(15) = 1350
10 feet wide: (100-20)(25-20) = 80(5) = 400

--------------------------------------------------------

Summary of my comments...

(1) You should know how to set up and solve this problem using formal algebra.

(2) You can get good mental exercise by solving the problem using logical reasoning and mental arithmetic.

(3) Often the formal algebraic solution is MUCH harder than the informal method.