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

a. What is the greatest rectangular area she can frame with the border?

b. What are the dimensions of the rectangle with greatest area?
c. If each dimension of the above rectangle were increased by 2 yards, what would be the dimensions and area of the new rectangle?

d. How much more is the new area?

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.

d. corresponding side lengths in similar triangles are proportional in length.

step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.