SOLUTION: A rectangular garden is to be surrounded by a walkway of constant width. The gardens dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What mus

Question 26465: A rectangular garden is to be surrounded by a walkway of constant width. The gardens dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What must be the width of the walkway to the nearest thousandth?
I am unable to figure this problem out at all. If you could please help me I would greatly appreciate it!! Thank you!!
Found 2 solutions by Earlsdon, cleomenius:
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Let the width of the walkway = x ft.
The total area, A, of the garden plus walkway can be expressed in terms of x as follows: A = L*W
Perform the indicated multiplication.
The total area is given as 8000 sq.ft., so:
Subtract 8000 from both sides of the equation.
Solve this quadratic equation for x. First factor out a 4.
Apply the zero products principle.
Solve using the quadratic formula:

= -38.86 Discard this negative solution.
= 3.86 ft.
The walkway is 3.86 feet wide.
Answer by cleomenius(959) (Show Source): You can put this solution on YOUR website!
The area is given by (x + 30) (x + 40) = 1800
x^2 + 70x + 1200 = 1800
x^2 + 70x -600 = 0

Next we use the quadratic equation.
-70 +,- Square root 4900 + 2400 / 2
= 7.720 (disregard the negative answer)
The areas are 37.720 * 47.720, which resolves to roughly ~1800 as a check.
Therefore, the width of the walkway is ~ 7.720

Cleomenius.

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