^{2}

^{2}+42W

^{2 }is negative. That makes the vertex the maximum for which we are searching.

^{2}+ bW + c = 0

^{2}

If you enclose a rectangle garden using a side of a building as one side of the rectangle, what are the dimensions of the garden if it is to be the maximum area that you can enclose with 42 feet of the fence?

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L = building side

W = non-building side

P = 2W + L = 42 (note only one L because the other L is the building itself)

Solve for L:

L = 42 - 2W

Area = l*w

Area = (42-2W)W = 42W - 2W^{2}

Let area be y, so y = -2W^{2} +42W

Note this is a parabola pointing down because the coefficient of the W^{2 }is negative. That makes the vertex the maximum for which we are searching.

Vertex of this parabola is at W=-b/2a, if the quadratic is aW^{2} + bW + c = 0

a = -2

b = 42

W = -42/(2*-2) = -42/-4 = 10.5

W = 10.5

L = 42-2(10.5) = 42-21 = 21

Area = L*W = 21 * 10.5

A = 220.5 ft^{2}

given a perimeter, you want to make a rectangle with length and width that maximizes the area of the rectangle

you maximize the area of the rectangle by making the length and the width the same, in other words you make a square, which is a special rectangle

for example if the perimeter is 40 units, make a square 40/4=10 units by 10 units; area equals 100 square units; a 9 by 11 rectangle gives us only 99, an 8 by 12 gives us only 96 and so on

in the problem we have 42 feet and if we divide 42 by 4 we get 42/4=10.5, so we would have a square

that is 10.5 by 10.5; 10.5x10.5=110.25, a 10 by 11 gives us only 110

however, we only have three sides of the rectangle

let the two widths be 10.5 each and the length be 42-2*10.5=42-21=21

10.5*21=220.5 sq ft

suppose we had 24 for the perimeter, 24/4=6

the widths and length would be 6,6, and 12 and the area would be 6*12=72 sq units

7,7 and 10 would only give us 70

5,5, and 14 would only give us 70

4,4, and 16 gives us 64

8,8, and 8 gives us 64

The area of a rectangle is length (l) times width (w).

You are using 42 feet of fence for 3 sides of the rectangle,

let's say 2 lengths and one width with the other width being the building.

So for A = lw

use

w = 42 - 2l

substitute

A = l (42 - 2l) = 42l - 2l^2

You can take the derivative of the area with respect to l and set equal to 0 to get l for maximum area. But since this is algebra you can also prepare a chart and graph to find maximum area.

dA/dl = 42 - 4l

l = 10.5

w = 21

OR

l w A

1 40 40

2 38 76

3 30 90

4 34 136

5 32 160

6 36 180

10 22 220

11 20 220

12 18 216

20 2 40

20 2 40

You can draw a graph of l vs A or w vs A and then get the point of graph for largest area which you can see is between l = 10 and l = 11.

Hi Erica, Normally the perimeter of a garden would be twice the length plus twice the width or P = 2L + 2W. But we know that the side of a building is going to be used as one side of the rectangle and we know that the other 3 sides added together = 42 feet. We also know that L + 2W = 42. Here are some options W = 14 L = 14 Area = 14*14 = 196.

W = 13 L = 16 Area = 13*16 = 208.

W = 12 L = 18 Area = 12*18 = 216.

W = 11 L = 20 Area = 11*20 = 220.

W = 10.5 L = 21 Area = 10.5*21 = 220.5.

W = 10 L = 22 Area = 10*22 = 220.

W = 9 L = 24 Area = 9*24 = 216.

W = 8 L = 26 Area = 8*26 = 208.

As you can see from the table the maximum area can be obtained by using 21 feet of the building as one side. So the dimensions for maximum area would be 10.5 feet by 21 feet.

Hey Erica -- box area is (42-2w)w or 42w-2w^2 ... we wish to find w to maximize A ...

try easiest widths: w=10 ... (22)10 or 220 ... w=11 ... (20)11 or 220 ... looks like peak

area is right in middle w=10.5 ... (21)10.5 = 210 +10.5 = **220.5 sq.ft. ...
**Regards :)

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## Comments

^{2}, basically. The UNITS are square feet.^{2}and mine was 220.5 sq feet. same thing.... my brain is fried from doing math all day. Sorry and thank you for your help. lol