Finding maximum/minimum area

Question

A square or a rectangle, whose length is 3 times greater than width, or both figures, are being wrapped up with a 24 m rope. What is the largest and the smallest area you can wrap up with the rope?

Answer 1

36 m^2 is the largest area, using just the square

about 15.44 m^2 is the smallest, using both figures

L=3W for the rectangle

24 = 2(3W + W) =8W

W =24/8 = 3 m

L= 9 m

Area of the rectangle = 27 m^2

4 sides of a square = 24 with each side =24/4 = 6. Area 36 m^2 = maximum area

for both figures

Area = ((24-8W)/4)^2 + 3W^2 =

=(6+2W)^2 + 3W^2

= 36+ 24W + 7W^2

take the derivative and set = 0, solve for W:

A' = 14W-24=0

W = 24/14 = 12/7 m = about wide for the rectangle

L=3W = 36/7 m long for the rectangle

area for the rectangle = 12/7 x 36/7 = about 8.82

4 sides of the square = 24-8W = 24-96/7 = 72/7

square side = 72/28= 18/7

area for the square =(18/7)^2 = 324/49 = about 6.61 m^2

+

total area for both = 8.82 + 6.62 = 15.44 m^2 = smallest area using both figures

the equation for the total area is a parabola opening upward, with vertex as the minimum point, its y coordinate as the minimum area.