Problem 1
L = length
W = width
P = perimeter of the rectangle
P = 2L+2W
P = 20
2L+2W = 20
L+W = 10 after dividing everything by 2.
L = 10-W
area = length*width
A = L*W
A = (10-W)*W
A = -W^2 + 10W
Let x replace W for a temporary basis.
If we graphed y = -x^2+10x, then we'd see the highest point occurs at (5, 25). This means a width of 5 meters leads to the largest area of 25 square meters.
The length would be L = 10-W = 10-5 = 5 meters as well.
It turns out that a 5 meter by 5 meter square is the best choice when trying to max out this rectangle's area.
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Problem 2
The results we got from the previous section was that a 5 by 5 square will max out the area.
If we allowed non-whole numbers into the mix, then it wouldn't change the previous answer. The best choice is still that 5 by 5 square.
The answer only would change if the result of the previous section was some non-integer value.