need help solving story problem

Question

a man wishes to put a around a rectangular field and then subdivide the field into 3 smaller rectangular plots by placing two fences parallel to one of the sides.  If he only has 3000 feet of fencing, what dimensions will give the maxium area?

Answer 1

draw a rectangle and divide it into the 3 smaller rectangles

in reality you have 2 lengths and 4 widths (1 width on either end and the 2 parallel fences inside the rectangle)

you have 3000 feet of fencing

how are the 2 lengths and the 4 widths related ?

the lengths and the widths take up all of the 3000 feet of fencing

3000=2l+4w is the relationship of the fencing to the perimeter and the two fences inside the rectangle

area=length*width

A=l*w

3000=2l+4w

divide both sides by 2

1500=l+2w

1500-2w=l

A=l*w

substitute

A=w*(1500-2w)

A=1500w-2w²

not knowing what level of math you are in, we'll solve the problem in two ways

set A=0

0=1500w-2w²

this is a parabola that opens down and its vertex, (h,k) is the maximum point

h is the "maximizing value" and k is the maximum area

0=1500w-2w²

using -b/2a

-1500/2(-2)=-1500/-4=1500/4=375 feet for "h" and for the width w

3000=2l+4w

3000=2l+4(375)

3000=2l+1500

3000-1500=2l

1500=2l

l=1500/2

l=750 feet for the length l

you have a rectangle 375 feet by 750 feet that will give the maximum area of 375*750=281,250 square feet which would be "k" in the vertex of the parabola

check:3000=750*2+375*4=1500+1500=3000 feet

if you know derivatives, take the first derivative of A

A'=1500-2(2)w

set A=0

0=1500-4w

4w=1500

w=1500/4

w=375 feet again and go from here