Algebra 2!!?

Question

I do not how to resolve this word problem!!!

"A farmer has 64 yards of fencing and wants to create a rectangular enclosure for his animals. What is the enclosure with the greatest area??"

Do you have to find two factors of 64 or what??

Answer 1

A=2L+2W

2L=2W

A=64

2L+2L=64
4L=64
L=64

16*16=256 yds^2

Answer 2

No, you want to maximize area...

So, you know for a rectangle:

length X width = Area

you know the total perimeter length must be: 64 yds...

and the equation is:

2*L + 2*W = 64 yds

We know then, L = 32 - W

Plug this into the Area equation:

A = (32-W)*W = 32*W-W^2

dA/dW = 0 = 32-2*W so, W must be 16.

Therefore, 2*(16) + 2*L = 64...

L = 16...

SO, W=16 and L=16... therefore the max area is

A = 16*16 = 256 yd^2

Or just by common sense, the largest area a rectangle can have is if it is a perfect square. So the sides must be equal.

I used calculus to solve this, sorry didn't notice the algebra 2 title.

Answer 3

Nevergive's answer is correct, but I'm sure you're expected todo this:

Let the length of the enclosure be x units, then the width is
32 - x (subtract length from half the perimeter of the rect)

Now the area is length * width
which = x(32 - x)

You then use either calculus or properties of quadratic expressions** to show the maximum value of the area is obtained when x = 16, so it's 256 sq ft.

**[If you graph y = x(32-x) you get a parabola which cuts the x axis at 0 and 32, therefore its axis of symmetry is at x = 16 so that's where the vertex is]

Answer 4

You need to maximize area, yet keep the perimeter the same

2x + 2y = 64, maximize xy

As the person above me realizes, the square is the optimal shape for maximizing area in a rectangle
so all sides are the same
x=y
that gives us 4x = 64
x=16

Answer 5

16*16=256

Answer 6

That sounds suspiciously like a problem I did in a few seconds, back when I was in Pre-K.