Finding Maximum Area

Question

What is the maximum area, in square feet, for the garden if 44 feet of fencing are used?

Answer 1

Hi Tyler,

You asked, "What is the maximum area, in square feet, for the garden if 44 feet of fencing are used?"

The 44 ft. of fencing are the garden's perimeter. We can use that to find possible dimensions of a rectangular garden.

Perimeter = 2•length + 2•width

If we think about this another way l + w = 1/2P

1/2P=22 ft.

What are all the possible length and width combinations that could make 22 ft?

length width

1ft. 21 ft.

2 ft. 20 ft.

3 ft. 19 ft.

4 ft. 18 ft.

5 ft. 17 ft.

6 ft. 16 ft.

7 ft. 15 ft.

8 ft. 14 ft.

9 ft. 13 ft.

10 ft. 12 ft.

11 ft. 11 ft.

If we keep going after this, we would just start repeated side lengths. Now that we have all the possible dimensions for a rectangle with a perimeter of 44, what are the possible areas? Well, let's complete the table.

length width area

1ft. 21 ft. 21 ft.²

2 ft. 20 ft. 40 ft.²

3 ft. 19 ft. 57 ft.²

4 ft. 18 ft. 72 ft.²

5 ft. 17 ft. 85 ft.²

6 ft. 16 ft. 96 ft.²

7 ft. 15 ft. 105 ft.²

8 ft. 14 ft. 112 ft.²

9 ft. 13 ft. 117 ft.²

10 ft. 12 ft. 120 ft.²

11 ft. 11 ft. 121 ft.²

From all of the above options, which dimensions give you a rectangle with the greatest area? Do you see a trend in the above chart? What do you think the rule is about rectangular dimensions and maximizing area?

I hope you found that answer helpful! If so, please upvote it.

Answer 2

The idea is to solve for one of the variables in the perimeter equation and then plug the value into the area equation.

For perimeter:

2L + 2W = 44

2L = 44 - 2W

L = (44-2W)/2

L = 22 - W

Since area (A) = L x W

A = (22 - W)W

A = 22W - W²

The new function is a parabola -- you can solve for the max by finding the roots (or zeroes). If we factor, we have 2 equations that we can set to equal 0.

22 - W = 0 and W = 0

By solving, we know that the parabola has zeroes when W = 0 and when W = 22. The maximum is located at the midpoint, W = 11.

Plugging W into the perimeter equation, to find L:

44 = 2L + 2(11)

22 = 2L

L =11

Therefore the max area is 11ft × 11ft or 121 ft²

Answer 3

Let us say that the garden has a rectangular shape.

With 44 feet of fencing the perimeter of the rectangle will be 44 feet.

The perimeter of a rectangle = 2 * length (l) + 2 width (w)

So 2l + 2w = 44 (this is your constraint equation)

l + w = 22 (dividing both sides of the equation by 2)

l = 22 - w (solving for length)

The area of a rectangle (A) = l * w (this is your optimization equation)

= (22 - w) * w (use above equation for l to put equation in one variable)

= 22*w - w**2 (separate terms)

Now we compute the derivative (A') = 22 - 2*w

Now set equal to zero and solve for w: 22 - 2*w = 0

w = 11

Since w = 11 and l = 22 - w, l = 11, so the shape that maximizes the area is a square.

The maximum area = l * w = 11 * 11 = 121 square feet.

Answer 4

If 44 feet of fencing are used, then the perimeter of the garden is 44 feet.

Let L = length (ft) and W = width (ft).

2L + 2W = 44. (four sides to a rectangle, two of each)

Divide by 2 throughout.

L + W = 22

Let's check the areas of the garden using lengths and widths that create a perimeter of 44 feet.

L W P(ft) A(ft²)

1 21 44 21

2 20 44 40

3 19 44 57

4 18 44 72

5 17 44 85

6 16 44 96

7 15 44 105

8 14 44 112

9 13 44 117

10 12 44 120

11 11 44 132

12 10 44 120

So, you can see the area started to decrease once we passed L=11 and W=11. This is because a square has the maximum area of any rectangle. If another problem like this comes up, save some time, and divide the perimeter by 4. If it is a round number, you have a square and that is your answer. If it's a decimal, try the round numbers near it on either side and see where the maximum area lies.

Answer 5

If the shape is not specified, the area is maximum when it is a circle

Forming a circle: circumference is 44 feet

2 * pi * r = 44

r = 7

Area = pi * r² = 154.06

If the shape is specified, the area is maximum when it is a square

Forming a square: each side will be 11 feet

Area = 11 * 11 = 121 feet²