3. The backyard of a home is a rectangle 25m by 30m. A garden of uniform width is to be built within the yard around the edge leaving a rectangle of grass inside. The area of the
grass should be the same as the area of the garden. What is the width of the garden?
Draw a diagram with your solution.​

Part A

The width of the garden is approximately 3.987 meters

Part B

Please find attached the drawing of the solution

Step-by-step explanation:

Part A

The given parameters are;

The dimensions of the backyard of the home = 25 m by 30 m

The width of the garden to be built in the yard = Uniform width

The shape of the grass left inside = Rectangular

The area of the grass at the center = The area of the garden

Let 'x' represent the width of the garden, we have;

The length of the rectangular grass area at the center, l = 30 - 2·x

The width of the rectangular area of grass at the center, w = 25 - 2·x

The area of the rectangular backyard, A = 30 m × 25 m = 750 m²

The area of the rectangular backyard, A = (The area of the garden) + (The area of the rectangle of grass inside)

The area of the rectangle of grass, GA = (30 - 2·x)·(25 - 2·x) = The area of the garden

The area of the rectangular backyard, A = 750 = (30 - 2·x)·(25 - 2·x) + (30 - 2·x)·(25 - 2·x)  = 2 × (30 - 2·x)·(25 - 2·x) = 8·x² - 220·x + 1,500

∴ 750 = 8·x² - 220·x + 1,500

8·x² - 220·x + 1,500 - 750 = 0

8·x² - 220·x + 750 = 0

x = (220 ± √(220² - 4 × 8 × 750))/(2 × 8)

x ≈ 23.513, or x = 3.987

When x = 25.513, the width of the rectangle of grass inside, w = 25 - 2 × 23.513 = -22.026, which is not a natural (physically possible)

Therefore, the possible width of the garden, x ≈ 3.987

Part B

The drawing of the solution created with MS Visio is attached

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)

step-by-step explanation: