Arthur D. answered • 10/14/17

Effective Mathematics Tutor

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-2w

^{2}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

Arthur D.

tutor

you asked for the area, A=l*w for a rectangle

Area=length times width (l*w)

the length comes from P=l+2w where P (perimeter) is the amount of fence 32 feet, thus 32=l+2w

32=l+2w means l=32-2w

the length is 32-2w (not 32w-2w) and the width is w

A=length* width

A=(32-2w)*w

use the distributive property

A=32w-2w

^{2}0=32w-2w

^{2}
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10/15/17

Katie R.

10/14/17