Question

Mrs. Del Pup bought 20 yards of border for her classroom bulletin board.

Answer 1

A rectangle with dimensions $x$ and $y$ has perimeter 20 if

$2(x+y)=20 \iff x+y=10$

and we can deduce

$y=10-x$

So, the dimensions of the rectangle must be $x$ and $10-x$, 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.

$x(10-x) = -x^2+10x$

The function $f(x)=-x^2+10x$ 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

$x=\dfrac{-b}{2a}=\dfrac{-10}{-2}=5$

And it yields an area of

$f(5)=-25+50=25$

(a) So, she can frame a maximum area of 25 squared yards.

(b)The dimensions of the rectangle with greatest area are $x=5$ and $y=10-x=10-5=5$, 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 $7^2=49$ squared yards. So, the new area is $49-25=26$ squared yeards more than the old one.