A rectangle with dimensions and
has perimeter 20 if
and we can deduce
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.
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
And it yields an area of
(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.