Question

I need some help with this question. Could you please help and explain?

Answer 1

Let's take a look at what's going on:

We know the perimeter of the playground (the sum of the lenght of its sides) has to be 460ft. Thereby,

`\begin{gathered} x+y+x+y=460 \\ \rightarrow2x+2y=460 \\ (\text{Equation 1)} \end{gathered}`

And the area has to be at least 8600 square ft. Remember that the are of a rectangle is the multiplication between the lenght of its base and the lenght of its height. Therefore,

`\begin{gathered} xy\ge8600 \\ \text{ (Equation 2)} \end{gathered}`

Now, let's clear x from (Equation 1) and substitute in (Equation 2). This will be our Algebraic method

`2x+2y=460\rightarrow2x=460-2y\rightarrow x=230-y`
`\begin{gathered} xy\ge8600\rightarrow(230-y)y\ge8600 \\ \rightarrow230y-y^2\ge8600\rightarrow0\ge y^2-230y+8600 \\ \rightarrow y^2-230y+8600\le0 \end{gathered}`

We get a quadratic inequality. We'll solve it as following:

0. Complete the square:

`\begin{gathered} y^2-230y+(115^2)+8600\le115^2 \\ \rightarrow(y-115)^2+8600\le115^2 \end{gathered}`

2. Simplify:

`\begin{gathered} (y-115)^2+8600\le115^2 \\ \rightarrow(y-115)^2\le115^2-8600 \\ \rightarrow(y-115)^2\le4625 \\ \end{gathered}`

Remember that

`a^2\le b\rightarrow-\sqrt[]{b}\le a\le√(b)`

3. Use this property:

`\begin{gathered} (y-115)^2\le4625 \\ \rightarrow-\sqrt[]{4625}\le y-115\le\sqrt[]{4625} \end{gathered}`

4. Add 115:

`-\sqrt[]{4625}+115\le y-115+115\le\sqrt[]{4625}+115`

5. Solve:

`47\le y\le183`

Now we know that the height can be between 115ft and 183ft. We've already cleared x, so let's substitude for both the values we've calculated:

`\begin{gathered} x=230-y \\ \rightarrow x=230-47\rightarrow x=183 \\ \rightarrow x=230-183\rightarrow x=47 \end{gathered}`

Thereby, we've calculated the lenght (x) of the playground has to be between 47ft and 183ft