Question 1187640: A rectangular playground has a walkway around it with a width of x-2. The garden is 36 m by 20 m.
Find an expression for the combined area of the playground and the walkway around it
Found 2 solutions by josgarithmetic, Theo:
Answer by josgarithmetic(39630)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! the length becomes 36 + 2 * (x - 2) which becomes 36 + 2x - 4 which becomes 32 + 2x.
the width becomes 20 + 2 * (x-2) which becomes 20 + 2x - 4 which becomes 16 + 2x.
the new dimensions are (32 + 2x) for the length and 16 + 2x for the width.
the combined area is (32 + 2x) * (16 + 2x) = 4x^2 + 96x + 512.
your equation for the combined area will be 4x^2 + 96x + 512 or (32 + 2x) * (16 + 2x).
note that x must be greater than 2, otherwise the combined area will be less than the original area.
the original area is 36 * 20 = 720
when x = 2, the combined area will be (32 + 4) * (16 + 4) = 36 * 20 = 720.
that's breakeven point.
for the path to have any width, x has to be greater than 2.
as a practical matter, if you want the path to be at least 2 feet wide, then x should be greater than or equal to 3.
the following graph shows the relationship.
as you can see, when x = 3, the combined area is 836 square feet.
that's because 32 + 2 * 3 = 38 and 16 + 2 * 3 = 22 and 38 * 22 = 836 square feet.
here's a diagram of what this would look like:
when x = 3, the value of A in the diagram is x - 2 which becomes 3 - 2 which becomes 1.
when x = 3, the value of A is 1.
the length is 36 + 2A = 38.
the width is 20 + 2A = 22.
the area is 38 * 22 = 836.
that's the same value on the graph when x = 3.
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