1. A man has 22 feet by 26 feet rectangular lot that he will use for planting. He wants to build a brick of walkway of uniform width on the border of the lot. If the man wants to have 396 square feet of ground left for planting, how wide should he build the walkway?

2. Based from Problem 1. The man buys enough brick to build 160 square feet of walkway. Is this sufficient to build the walkway mentioned in Problem 1? If not, how wide can she build he walkway with these bricks?

3. Refer to Problem 1. The man has 50 feet of fencing that he will use to enclose an area for flowering plants. Find the dimensions of the largest area he can fence. Find the largest possible rectangular area he can enclose.

(22-2w)(26-2w) = 396

w = 2

24*26-396 = 228
No, not enough brick
22*26 - (22-2w)(26-2w) = 160
w = 1.8

Not sure how #3 relates to #1
as with all such problems, divide the fencing evenly between the lengths and the widths. That means that a square 50/4 feet on a side will provide the maximum area. To see this, note that if the area has dimensions x and y,
2x+2y = 50
y = 25-x
The area is x*y = x(25-x) = 25x-x^2
This is just a parabola with vertex at x=25/2, as described above.

1. To find the width of the walkway, we can subtract the area of the rectangular lot and the desired remaining ground area.

First, let's calculate the area of the rectangular lot:

Area of the rectangular lot = Length * Width = 22 feet * 26 feet = 572 square feet

Now, subtract the desired remaining ground area from the area of the rectangular lot to get the area occupied by the walkway:

Area of the walkway = Area of the rectangular lot - Desired remaining ground area = 572 square feet - 396 square feet = 176 square feet

Since the walkway is rectangular, we can represent its length as (22 + 2x) feet (where x represents the width of the walkway).

The width of the walkway can be represented as (26 + 2x) feet.

The area of the walkway can also be calculated using these dimensions:

Area of the walkway = Length * Width = (22 + 2x) feet * (26 + 2x) feet = 176 square feet

Now we can solve this quadratic equation for x by multiplying the binomials:

(22 + 2x) * (26 + 2x) = 176

Distribute and rearrange the equation:

4x^2 + 96x + 572 = 176

Bring all terms to one side:

4x^2 + 96x + 572 - 176 = 0

Simplify:

4x^2 + 96x + 396 = 0

Divide the equation by 4 to simplify:

x^2 + 24x + 99 = 0

Now we have a quadratic equation that we can solve using factoring, completing the square, or using the quadratic formula. In this case, factoring or completing the square does not give us a nice solution, so we can use the quadratic formula:

x = (-24 ± √(24^2 - 4 * 1 * 99)) / (2 * 1)
x = (-24 ± √(576 - 396)) / 2
x = (-24 ± √180) / 2
x ≈ (-24 ± 13.416) / 2

x ≈ (-24 + 13.416) / 2 or x ≈ (-24 - 13.416) / 2

x ≈ -10.584/2 or x ≈ -37.416/2

Since the width cannot be negative in this context, we can discard the negative value.

Therefore, the width of the walkway should be approximately 5.292 feet.

2. To determine if the man has bought enough brick to build the walkway mentioned in Problem 1, we need to calculate the area of the walkway using the width we found in the previous question.

Area of the walkway = (22 + 2x) feet * (26 + 2x) feet
Area of the walkway = (22 + 2*5.292) feet * (26 + 2*5.292) feet
Area of the walkway ≈ 204 square feet

Since the area of the walkway is 204 square feet, and the man has bought enough brick to build 160 square feet of walkway, it is not sufficient to build the walkway.

3. To find the dimensions of the largest area the man can fence using 50 feet of fencing, we need to determine the shape that maximizes the enclosed area. Since the man wants to enclose a rectangular area, the dimensions that maximize the area will be a square shape.

To find the largest possible rectangular area, we divide the total length of fencing by 4 (since a square has all sides equal).

Length of each side of the square = Total length of fencing / 4 = 50 feet / 4 = 12.5 feet

Therefore, the dimensions of the largest area he can fence is a square with each side measuring 12.5 feet.