A rectangle with dimensions [tex]x[/tex] and [tex]y[/tex] has perimeter 20 if
[tex]2(x+y)=20 \iff x+y=10[/tex]
and we can deduce
[tex]y=10-x[/tex]
So, the dimensions of the rectangle must be [tex]x[/tex] and [tex]10-x[/tex], 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.
[tex]x(10-x) = -x^2+10x[/tex]
The function [tex]f(x)=-x^2+10x[/tex] 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
[tex]x=\dfrac{-b}{2a}=\dfrac{-10}{-2}=5[/tex]
And it yields an area of
[tex]f(5)=-25+50=25[/tex]
(a) So, she can frame a maximum area of 25 squared yards.
(b)The dimensions of the rectangle with greatest area are [tex]x=5[/tex] and [tex]y=10-x=10-5=5[/tex], 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 [tex]7^2=49[/tex] squared yards. So, the new area is [tex]49-25=26[/tex] squared yeards more than the old one.
