Katherine S.
asked • 01/22/25Trying to explain this to my 3rd grader without algebra. No teacher help- so frustrated. This is a test question (we are reviewing what she got wrong) from Math Expressions unit 6. Do I just say make it a square and divide by 4, and then multiply the length x width? I have no idea how the teacher taught it and no idea how this child would explain how she got the answer.This question seems insane for 3rd grade!!!
A farmer has 80 feet of fencing to fence four sides of a rectangular garden. What is the greatest area of garden that the farmer can fence? Explain how you know.
Please help me- I want her to understand this, but I need a simpler explanation on a 3rd grade level.
I would let her discover the answer. Let her start with a rectangle and continue to make shorter side of rectangle smaller while making larger size smaller and find area.
Ex
5x35 rectangle has area of 175
10x30 rectangle has area of 300
15x25 rectangle has area of 375
20x20 rectangle has area of 400
25x15 rectangle has area of 375
Etc
This kind of problem is not given within a "normal" curriculum for every 3rd grader. But I've seen this kind of level of problem in advanced programs and as an optional/advanced/fun sections of some good textbooks.
If "solve without Algebra" was a requirement, I would go for the following explanation for a 3rd grader. Actually, as you will see, I will use (hidden) Algebra, without formally introducing variables.
--------------------------------------------------------------------------------------------
A. If length and width were EQUAL to one another, then they would be equal to
80 / 4 = 20, then
the Area would be 20 * 20 = 400 (case A).
B. What if length and width were NOT equal?
The total of length + width is half of the perimeter, or 40.
If the longer side is more than 20 by some value, then the shorter side must be less than 20 by the same value. We will call this same value "difference" (compared to 20).
So length will be (20 + difference), and
width will be (20 - difference), and
the area will be (20 + difference) * (20 - difference),
then (ether with the full calculation or, if the 3rd grader was ready to use the "difference of squares" formula) it comes to
Area = 20 * 20 - difference * difference (case B),
which means that, if there was any 'difference' between length and width, we would be
subtracting the value of (difference * difference) from 400.
In case A, when all sides of the rectangle garden were equal, the area of the rectangle garden was 400.
So 400 if the greatest possible area of the rectangle garden. Note: when the rectangle is a square!
Answer: The greatest area of garden that the farmer can fence is 400 sq. ft.
Arthur D. answered • 01/22/25
Forty Year Educator: Classroom, Summer School, Substitute, Tutor
Hi Katherine,
I would begin by drawing a diagram. Draw a rectangle and explain that the opposite sides of a rectangle have the same length or measure and that one horizontal side is called the length and one vertical side is called the width. This is not beyond the comprehension of a third grader. Having a diagram to look at helps in understanding and a third grader can clearly see that the opposite sides of a rectangle, if drawn neatly with a ruler, appear to have the same length. Your daughter can also measure the opposite sides herself with the ruler. Now explain that the lengths of all of the four sides added together must equal 80 feet because that is how much fencing the farmer has. Again, stress that the opposite sides have the same length. You can also explain that one length and one width added together must equal 40 feet, half of 80 feet. Now you have to use trial and error. Choose two numbers, one to represent the length and one to represent the width, that add up to 40 feet. Because 80 is a multiple of 10 and a multiple of 5, you can suggest that she use multiples of 5 or multiples of 10. Now see what your daughter choses for numbers and guide her and correct her if she makes any arithmetic mistakes. From here on you increase one number by 5 (or 10) and decrease the other number by 5 (or 10). You would now have to explain how to find the area of a rectangle. Area=length x width or A=L x W. Your daughter will soon discover a pattern whereby as the smaller number increases and the larger number decreases, the smaller number now becomes the larger number and the larger number now becomes the smaller number and you now have repetition of numbers.
You find the area each time you change numbers until your daughter realizes that 20 x 20 will give the largest area. This is true in mathematics; a square has the largest area. Doing the problem this way gives your daughter practice in adding and multiplying as well as patterns.
5 x 35
10 x 30
15 x 25
20 x 20
25 x 15
30 x 10
35 x 5
After you come to the conclusion that 20 x 20 gives the largest area, you can try other numbers that are not multiples of either 5 or 10. You could use 8 and 32, 12 and 28, 16 and 24, and whatever other numbers you like.
The algebra for this type of problem is as follows:
P=2l+2w and A=lw
80=2l+2w, 40=l+w, 40-l=w (or 40-w=l)
A=lw, A=l(40-l), A=40l -l^2, A=-l^2+40l
The maximizing point is h=-b/2a where a=the coefficient of l^2 which is -1 and b is the coefficient of l which is 40. Therefore h=-40/2(-1), h=-40/(-2), h=20
A=lw, A=20(20)=400 square feet. Each side of the rectangle must be 20 feet so you have a 20 by 20 square.
PASINDU P. answered • 01/30/25
A good tutor makes you feel comfortable
I totally get how frustrating this can be! Here’s a simple way to explain it without algebra, using hands-on thinking that a 3rd grader can understand.
**Step 1: Start with the Perimeter Concept**
Tell your child:
*"We have 80 feet of fence, and we need to use all of it to make a rectangle. That means the total length of all four sides must add up to 80 feet."*
**Step 2: Try Different Rectangles**
Instead of jumping to a square, encourage your child to think about different rectangles they can make with 80 feet of fencing. For example:
- If the garden is **10 feet wide**, then the two widths take up **10 + 10 = 20 feet**. That leaves **60 feet** for the two lengths, so each length would be **30 feet** (since 60 ÷ 2 = 30).
- If the garden is **20 feet wide**, then the two widths take up **20 + 20 = 40 feet**, leaving **40 feet** for the two lengths, so each length would be **20 feet** (since 40 ÷ 2 = 20).
**Step 3: Find the Area**
Explain:
*"The area is how much space is inside the garden. We find the area by multiplying the length times the width."*
- **10 x 30 = 300 square feet**
- **20 x 20 = 400 square feet**
**Step 4: Find the Biggest One**
Ask your child:
*"Which one is bigger, 300 or 400?"*
Since **400 is bigger**, the best choice is **a 20 ft by 20 ft square garden**.
**How to Explain Why a Square Works Best**
If they ask why, you can say:
*"A square spreads out the fencing evenly, so we get the most space inside. If we make the garden too long and skinny, we lose space."*
You can even draw different rectangles and count squares inside to show which one holds the most.
Would this explanation help your daughter understand? 😊
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![{\displaystyle {\begin{aligned}&{\text{Tools: graph paper (to draw rectangles easily).}}\\&{\text{Start with any rectangle satisfying the perimeter requirement, e.g., }}30\times 10.{\text{ How can }}\\&{\text{we change its shape and still keep the perimeter at }}80?{\text{ Shrinking its length, say, by }}5\\&\qquad 30\rightsquigarrow 25\\&{\text{would reduce the perimeter by }}5+5{\text{ (two segments are cut off). To keep the perimeter }}\\&{\text{unchanged, you need to increase the width by }}5,\\&\qquad 30\times 10\rightsquigarrow 25\times 15.\\&{\text{In general, to keep the perimeter, possible changes have the form}}\\&\qquad L\times W\rightsquigarrow (L-c)\times (W+c)\quad {\text{ or }}\quad L\times W\rightsquigarrow (L+c)\times (W-c),\\&{\text{but don't use variables in your explanation, stick to numerical examples.}}\\&{\text{To compare their areas, align the two rectangles at the lower left vertex. }}\\&{\text{They share a core rectangle of }}25\times 10,\\&\qquad {\begin{aligned}30\times 10&=(25\times 10)+(5\times 10)\\25\times 15&=(25\times 10)+(25\times 5)\end{aligned}}\\&{\text{The two 'flaps' have one side }}(5){\text{ in common, so their areas are easy to compare. }}\\&{\text{The flap with the longer side }}(25>10){\text{ has more area. Therefore, }}25\times 15{\text{ has more }}\\&{\text{area than }}30\times 10.\\&{\text{It's now not difficult to see that as long as }}L>W,{\text{ I can choose a }}c>0{\text{ such that }}\\&\qquad L-c>W+c\qquad {\text{[still longer than wide]}}\\&{\text{and the new rectangle has greater area (}}L-c>W{\text{). E.g., }}\\&\qquad 30-9>10+9.\\&21\times 19{\text{ has the greater area than }}30\times 10{\text{ (flap argument) but the gap has shrunk}}\\&{\text{from }}30-10=20{\text{ to }}21-19=2.{\text{ For the greatest area, choose }}c{\text{ such that}}\\&\qquad L-c=W+c\,.\\&{\text{If you started with }}30\times 10,{\text{ this will yield the square }}(30-10)\times (10+10).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d3c461a5f9da48397dd686b08813398f38809d9)