Question

A rectangular garden is 21 feet long and 14 feet wide. A brick walk way with uniform width is going to be constructed around the garden. The brick walk way will have an area of 330 square feet. What should be the width of the brick walk way? Type in your answer to the nearest thousandth. The width of the walk way should be?

Answer 1

The width of the brick walk way in the rectangular garden should be 1/2 feet

How to determine what the width of the brick walk way should be.

From the question, we have the following parameters that can be used in our computation:

Dimension of garden = 21 feet by 14 feet

Also, we have

Area of the walkway = 330 square feet

Let the width of the brick be x

So, we have the following dimenson

21 + 2x by 14 + 2x

The area is then calculated as

(21 + 2x) * (14 + 2x) = 330

Expand

4x² + 70x + 294 = 330

4x² + 70x + 294 - 330 = 0

4x² + 70x - 36 = 0

Divide through by 2

2x² + 35x - 18 = 0

Expand

2x² + 36x - x - 18 = 0

This gives

2x(x + 18) - 1(x + 18) = 0

So, we have

(2x - 1)(x + 18) = 0

When evaluated, we have

x = 1/2 and x = -18

Hence, the width is 1/2 because the width cannot be negative (i.e., we discard the value x = -18)

Answer 2

3.862 feet

Explanation:

Let the uniform width of the walkway = x ft

From the diagram above, the area of the walkway will be:

`2x(21+2x)+2(14x)`

The brick walkway will have an area of 330 square feet. Therefore:

`2x(21+2x)+2(14x)=330`

We simplify the equation in order to solve for x:

`\begin{gathered} 42x+4x^2+28x=330 \\ 4x^2+28x+42x-330=0 \\ 4x^2+70x-330=0 \end{gathered}`

Next, solve for x:

`\begin{gathered} 2(2x^2+35x-165)=0 \\ 2x^2+35x-165=0 \end{gathered}`

Using the quadratic formula, we solve for x:

`\begin{gathered} $$x=(-b\pm√(b^2-4ac) )/(2a)$$ \\ x=(-35\pm√(35^2-4(2)(-165)))/(2*2) \\ =(-35\pm√(1225-(-1320)))/(4) \\ =(-35\pm√(2545))/(4) \\ x=(-35+√(2545))/(4)\text{ or }x=(-35-√(2545))/(4) \\ x=3.862\text{ or }x=-21.362 \end{gathered}`

The width of the brick walkway should be 3.862 feet (correct to the nearest thousandth).