You can put this solution on YOUR website!
Your question seems a bit vague.
My interpretation is that you're given some amount of fencing (a fixed constant perimeter P) and your teacher wants you to prove that the largest rectangle area happens when it's a square of side length 0.25P
Recall that any square is a rectangle but not vice versa.
The formula for the perimeter of rectangle is
P = 2*L + 2*W
That can be rearranged to
W = 0.5P - L
The algebra shouldn't be too tricky.
Let me know if you have questions about this portion.
Replace 0.5 with 1/2 if you want.
Next I'll use x in place of L.
x = length
0.5P - x = width
area = length*width
A = x*(0.5P - x)
A = 0.5Px - x^2
A = -x^2 + 0.5Px
y = -x^2 + 0.5Px
This graphs an upside down parabola because the leading coefficient is negative. You can use a graphing tool like Desmos or GeoGebra (there are many others to choose from).
Try graphing an example function like y = -x^2+4x or y = -x^2+10x to see what I mean.
Anyways let's return back to y = -x^2 + 0.5Px
The highest point of this parabola is the vertex.
It's where the rectangular area maxes out.
The roots occur when y = 0
x*(0.5P - x) = 0
x = 0 or 0.5P - x = 0
x = 0 or x = 0.5P
If x = 0 or x = 0.5P, then the area is 0.
The midpoint of the roots is the x coordinate of the vertex due to the parabola's symmetry.
The midpoint of 0 and 0.5P is 0.25P
Therefore the x coordinate of the vertex is x = 0.25P
length = x = 0.25P
width = 0.5P - x = 0.5P - 0.25P = 0.25P
Both length and width are 0.25P
We have a square with side length 0.25P
This concludes the proof.
We have proven that given a fixed perimeter (P), the max rectangular area occurs when the length and width are 0.25P (i.e. when we have a square)
If you're confused about the wording "fixed perimeter" it just means you're given some amount of fencing and cannot change the number.
For instance if you're given P = 200 feet of fencing then 0.25*P = 0.25*200 = 50 feet is the side length of the square and the area is 50^2 = 2500 square feet. This is the largest rectangle area possible for this amount of fencing.
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