Is it possible to have a figure an area of 9 square units and a perimeter of 30? If not, why? (PS I don't think it is possible.)

Answer 1

LW=9
2L+2W=30

that is for a rectangle.

2*L/9+2L=30
20L=270
L=13.5, W=13.5/9=you do it.

Answer 2

sure maximum area for a given perimeter is a square.

A square with perimeter 30 has area 7.5^2 = 56.25

So, since 9 < 56.25, we can find such a rectangle.

If the width is x, then the length is 15-x, so

x(15-x) = 9
x^2 - 15x - 9 = 0
x = 3/2 (5 - √21) = 0.62614

the rectangle has area

= 9/4 (5-√21)(5+√21)
= 9/4 (25-21)
= 9/4 * 4
= 9

Answer 3

LW=9
2L+2W=30

that is for a rectangle.

2*L/9+2L=30
20L=270
L=13.5, W=13.5/9=you do it.

Answer 4

W=9/L, not W=L/9