I have a word problem. You might have to use algebra for solving.?

Question

A rectangular field will be constructed near the Dog River with one side of the fiedl along the edge of the river (where the river is straight). No fence is needed along the riverbank. There is fence material available that is 400 meters long.
a) Suppose that the length of the fence (parallel to the river) is 80 meters.
Find the width of the fence (perpendicular to the river) and compute the area enclosed by the fence.
b) Suppose that the width of the fence (perpendicular to the river ) is 100 meters. Find the length of the fence (parallel to the river) and compute the area enclosed by the fence.
c) What length (parallel to the river) and width (perpendicular to the river) of the fence can you choose so that the area of the field is maximized? What is the maximum area that can be enclosed within the field?

Answer 1

a) W + W + 80 = 400
2W = 320
W = 160 meters

Area = W * L = 80 * 160 = 12,800 sq. m

b) 2(100) + L = 400
L = 200 meters

Area = W * L = 100 * 200 = 20,000 sq. m


c) 2W + L = 400
Solve for one letter, eg. L
L = 400 - 2W

Now plug that into the Area formula:
A = W*L
A = W(400 - 2W)
A = 400W - 2W^2

To maximize, take the derivative and set to zero.
400 - 4W = 0
4W = 400
W = 100

So the area is maximized when you have sides of 100, 200, 100.
W (width perpendicular to the river) is 100 m)
L (length parallel to the river) is 200 m
Enclosed area (maximum) is 20,000 sq. meters.

Answer 2

Forget the Algebra, just think about it.

The river will act as a fence and need no material. That leaves 80 meters for the opposite and parallel side of the river. Since a rectangle has opposite sides parallel and congruent.
400-80 = 320 meters of materials left for the other two opposite and congruent sides. 320/2 = 160 meters for each side of the remaining rectangle.
The area enclosed would be A = lw; 80*160 = 12,800 square meters.

Use the same thought process for part b. You know the width of the fence is 100 meters. There are two sides congruent and parallel which would require 2 * 100 = 200 meters needed.
400 - 200 = 200 meters left for the remaining side opposite the river.
Then the area would be 100 * 200 = 20,000 square meters.

The area is maximized at 20,000 square meters. If you start reducing one of the sides and add it to the other side, calculating the area will result in a smaller number.

Answer 3

a) After taking out the 80m for the side parallel to the river, there are 320m left over, to be divided evenly among the other two sides. Therefore the width is 320/2 = 160m, and the area is 80*160 = 12,800 m^2.

b) If the width is 100m, then 200m is needed for the two sides perpendicular to the river, and there is 200m left over for the one side parallel to the river. Therefore that is the length, and the area is 200*100 = 20,000 m^2.

c) Let x be the length. Then the width is (400-x)/2, and the area is x(400-x)/2 = 200x - x^2/2. To find the value at which this is maximum (I'm assuming calculus is not allowed), you can rewrite this by completing the square:

-x^2/2 + 200x = -1/2(x^2 - 400x)
= -1/2(x - 200)^2 + 20000.

-1/2(x-200)^2 <= 0 for all x, so the maximum occurs when this part is zero. This is when x=200. Therefore the maximum area, from part (b), is 20,000 m^2, from length 200m and width 100m.

Answer 4

A. Always draw a picture with word problems, it really helps. You know that the perimeter of a rectangel is the sum of all the side so you know that the side parallel to the river is 80 meters. The problem stated that the river did not require fencing, so with 400 m of available fencing:

400m - 80m = 320 m for the other 2 sides

320/2 = 160m is the width of the fence.

B. Solve this the same way.

400 m - 2(100m) = 200 m

the formula for area of a rectangle is Area = length x width

Area = 100m x 200m = ?

C. hint try drawing the side opposite the river 75 feet, 100 feet, and 125 feet and see what size area you get.

Answer 5

First, establish an easy working formula for circumference and area.

Circumference = Length + 2*Width
Area = Length * Width

a) If the length is 80 meters and you have 400 to work with, then
Cirumference = 80 + 2*Width = 400
Width = 160 meters
Area = Length * Width = 80 * 160 = 12,800 square meters

b) Circumference = Length + 2*100 = 400
Length = 200
Area = 200*100 = 20,000 square meters

c) The maximum area under these specifications will require very basic calculus. (I'm sure you can do it another way, but this is easiest for me). Let's call length X and width Y

Circumference = X + 2*Y= 400
If you solve for X you see that X = 400 - 2*Y
Now you can plug this into the area formula:

Area = X*Y
Area = (400-2*Y)*Y
Area = -2*Y^2 + 400*Y

Now you take the derivative of the area with respect to variable Y and set it equal to zero to find it's maximums and minimums.

dArea/dY = -4*Y + 400 = 0
Y = Width = 100

And because Circumference = X + 2*Y= 400
X = 200

Area = 200*100 = 20,000

So in problem b) you see that this is actually the ideal situation.

Answer 6

a) w = (400-80)/2 = 320/2 = 160.
A=160*80 = 12,800

b)L=400-(2*100) = 200
A=200*100=20,000

c)A=w*L
2w+L=400, or L=400-2w.
So, substitute in L, getting A=w(400-2w) = 400w-2w^2.
Take the derviative and set to 0, giving you 400-4w=0,
or 400=4w, so w=100, and L=400-(2*100) = 200.

So, the width=100 and L=200, and maximum area= 100*200 = 20,000 .

Answer 7

2(width)+length = 400
(a) 2(width)+length = 400
2(width)+80 = 400
2(width) = 400-80 = 320
(width) = 320/2 =160
Area = width*length = 160*80 = 12800 sqm
(b) 2(100)+length = 400
200+length = 400
length = 400-200 = 200
Area = width*length = 100*200 = 20000 sqm
(c) Let w be the width and i be the length
then 2w+l = 400..eqn 1
and area = w*l..eqn 2
from eqn 1 l = 400-2w...eqn 3
substitute for l from eqn 3 into eqn 2
so area = w*(400-2w)
= 400w - 2w^2
for area to be maximum , derivative of area has to be =zero
derivative of area = 400 - 4w = 0
so 4w = 400 or w=100
eqn 1 says 2w+l = 400
substitute w = 100
then 2(100)+l = 400
so l = 200
area (max) = w*l = 100*200 = 20000 sqm

Answer 8

a) width of fence is (400-80)/2 = 160
area = 160*80=12800
b) length of fence is 400-2*100=200
area = 100*200 = 20000
c)let the width of the fence be x
length of fence is 400-2x
area, y = (400-2x)x
=400x - 2x^2
diffentiate wrt to x,
dy/dx = 400-4x
when dy/dx = 0
x = 100
Therefore, width of fence is 100, length of fence is 200, area is 20000