Question

A club swimming pool is 18 ft wide and 34 ft long. The clubmembers want an exposed aggregate border in a strip ofuniform width around the pool. They have enough material for348 ft. How wide can the strip be?The width of the border will beft.

Answer 1

A rough diagram of the pool:

We want a border around the rectangular pool that has uniform width. Let that width be "x". A rough diagram of this is shown below:

So, the perimeter (sum of all the sides) of the outer border can be written as:

$(18+x+x)+(34+x+x)+(18+x+x)+(34+x+x)$

This is equal to 348, since enough material for 348 feet is available. Thus, we can write:

$(18+x+x)+(34+x+x)+(18+x+x)+(34+x+x)=348$

Now, we can do a bit algebra and solve for "x". This is shown below:

$\begin{gathered} (18+x+x)+(34+x+x)+(18+x+x)+(34+x+x)=348 \\ 18+2x+34+2x+18+2x+34+2x=348 \\ 104+8x=348 \\ 8x=348-104 \\ 8x=244 \\ x=\frac{244}{8} \\ x=30.5 \end{gathered}$Answer30.5