A rectangle with dimensions
and
has perimeter 20 if
![2(x+y)=20 \iff x+y=10](https://tex.z-dn.net/?f=2%28x%2By%29%3D20%20%5Ciff%20x%2By%3D10)
and we can deduce
![y=10-x](https://tex.z-dn.net/?f=y%3D10-x)
So, the dimensions of the rectangle must be
and
, where x ranges from 0 to 10 (both extremes are excluded, otherwise you'd have a degenerate rectangle, which is actually a segment).
So, all the possible areas are given by the product of the dimensions, i.e.
![x(10-x) = -x^2+10x](https://tex.z-dn.net/?f=x%2810-x%29%20%3D%20-x%5E2%2B10x)
The function
represents a parabola, facing downwards, and thus it admits a maximum, which is its vertex.
In particular, the vertex of this parabola is given by
![x=\dfrac{-b}{2a}=\dfrac{-10}{-2}=5](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B-b%7D%7B2a%7D%3D%5Cdfrac%7B-10%7D%7B-2%7D%3D5)
And it yields an area of
![f(5)=-25+50=25](https://tex.z-dn.net/?f=f%285%29%3D-25%2B50%3D25)
(a) So, she can frame a maximum area of 25 squared yards.
(b)The dimensions of the rectangle with greatest area are
and
, so it's actually a square.
This is a well known theorem: if you fix the perimeter, the rectangle with the largest area is the square yielding that perimeter.
(c) If we increase both dimensions by 2 yards, our 5x5 square would become a 7x7 square...
(d) ...and the new area would be
squared yards. So, the new area is
squared yeards more than the old one.