Word Problem Help!

Question

A fourth-grade class decides to enclose a rectangular garden, using the side of the school as one side of the rectangle. What is the max area that the class can enclose using 32 ft of fence? What should the dimension of the garden be in order to yeild this area?

Answer 1

P=2l+2w but one length is not used because of the school

so P=l+2w

32=l+2w and l=32-2w

A=l*w

A=(32-2w)(w)

0=32w-2w²

this is a parabola that opens downward so there is a maximum at (h,k) where h is the "maximizing point" and k is the maximum area

use h=-b/2a

h=-32/-4

h=8 feet

you would ordinarily have a square that is 8 feet by 8 feet but one section of 8 feet is not used

put this 8 feet with the opposite 8 feet to have a rectangle whose measurements are 8 feet by 16 feet by 8 feet by "the school"

A=lw

A=16*8

A=128 square feet is the maximum area

if you had a rectangle 17 by 7.5 the area would be 17*7.5=127.5 square feet which is less

if you had a rectangle 15 by 8.5 the area would be 15*8.5=127.5 square feet again

14*9=126

18*7=126