enclose a rectangular areausing the creek as one side and fencing for the other three sides If there is 380 ft of fencing availble . what is the max possible?

Question

I don't know how to set up or solve. Someone help please :)

Answer 1

P=2l+2w

380=2l+2w

190=l+w

190-w=l

A=lw

A=(190-w)w

0=190w-w²

this is a parabola that opens downward so there is a maximum, which is the vertex

(h,k) is the vertex, where h is the maximizing value and k is the maximum area

0=190w-w²

h=-b/2a

h=-190/(-2)

h=95 feet

the creek does not need fencing so you have an extra 95 feet

ordinarily you would have a square but you don't need 95 feet of fencing

take the 95 feet that you don't need and add it to the opposite side

you have a rectangle that is 95 feet by 190 feet (95+95) by 95 feet by "the creek" (the fourth side that has no fencing)

A=lw

A=190*95

A=18,050 square feet is the maximum area

notice that the rectangle is actually two squares 95x95 put together side by side

Answer 2

Hello Caylee

 

The rectangle with the greatest area is a square.

 

To prove this, let's consider 8 feet of fencing to surround all 4 sides.  A square would be 2 feet on each side, which would give us an area of 4 sq ft (2x2).  A rectangle of length 3 and width 1 would only result in an area of 3 sq ft (3x1).

 

 

We have 380 ft of fencing which will enclose an area with a creek on one side.  This means we only need the fencing for 3 of the 4 sides.

 

Since we want a square, each side will be

380/3 = 126 2/3 feet

 

Now if each side is 126 2/3 feet and Area = side ²

then Area = (126.6666)² = 16044.44 square feet