Word problems about quadratic functions

Question

Her father wants to make a small rectangular fenced in area for his children to play in the backyard He has 60 meters of fencing material to cover three sides of the area with the house acting as the fourth side . What dimension should the father use for each side of the fence maximize the area for his children’s place

Answer 1

2w + L = 60

Area = length x width

= L w

= w ( 60 - 2w)

60w - 2w^2 is a function of w

The zeros area 0 = 60w - 2w^2

0 = 2w (30 - w)

w = 0 or w=30

The max occurs at the average of the zeros

which is 15.

60*15 - 2*15^2 =

900 - 2 * 225 =

900 - 450 =

450

THe max area is achieved by rectangle 15 by 30