Question

A rectangular park is 60 meters wide and 105 meters long. Give the length and width of another rectangular park that has the same perimeter but a smaller area.

Answer 1

First we find the area of the first park

`\begin{gathered} A=w* l \\ A=60*105 \\ A=6300 \end{gathered}`

area is 6300 square meters

now find the perimeter

`\begin{gathered} P=2w+2l \\ P=2(60)+2(105) \\ P=330 \end{gathered}`

perimeter is 330 meters

now we need to write equations to find the measures of the another park and we can write from the statements

has the same perimeter

`2w+2l=330`

but a smaller area

then we choose an area smaller than 6300, for example 6000

`w* l=6000`

now we have two equations and two unknows

`\begin{gathered} 2w+2l=330 \\ w* l=6000 \end{gathered}`

then we can solve a unknow from one equation and replace on the other

I will solve w from the first equation and replace on second

`\begin{gathered} 2w=330-2l \\ w=(330-2l)/(2) \\ \\ w=165-l \end{gathered}`
`\begin{gathered} w* l=6000 \\ (165-l)* l=6000 \\ 165l-l^2=6000 \end{gathered}`

rewrite the equation

`l^2-165l+6000=0`

and use quadratic formula to solve L

`\begin{gathered} l=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \\ l=\frac{-(-165)\pm\sqrt[]{(-165)^2-4(1)(6000)}}{2(1)} \\ \\ l=\frac{165\pm\sqrt[]{27225-24000}}{2} \\ \\ l=\frac{165\pm\sqrt[]{3225}}{2} \end{gathered}`

then we have two values for l

`\begin{gathered} l_1=\frac{165+\sqrt[]{3225}}{2}=110.9 \\ \\ l_2=\frac{165-\sqrt[]{3225}}{2}=54.1 \end{gathered}`

we can take any value because both are positive and replace on any equation to find w

I will replace l=110.9 to find w

`\begin{gathered} w* l=6000 \\ w*110.9=6000 \\ w=(6000)/(110.9) \\ \\ w=54.1 \end{gathered}`

Finally the length and wifth of the other rectangle patks are

`\begin{gathered} l=110.9 \\ w=54.1 \end{gathered}`

meters