The area is equal to length times width.
![A=lw](https://img.qammunity.org/2019/formulas/mathematics/middle-school/jqxsv1ebncbkrat5p919iu4sbs13jmg52u.png)
The perimeter is equal to twice the sum of the length and the width.
![p=2l+2w](https://img.qammunity.org/2019/formulas/mathematics/high-school/m8zvmw9etca9fegsouayvn3t3b1tm9yodp.png)
We know that the perimeter is 450 meters, the length is
![450-2x](https://img.qammunity.org/2019/formulas/mathematics/college/2eubuzyjuslpjdrolm3j9oalxo26qpljwy.png)
meters, and the width is
![x](https://img.qammunity.org/2019/formulas/mathematics/college/lhtxftojjkzsmo3o2h4ilq8naohracejui.png)
meters.
To maximize the area, we find the global maximum of the function
![a(x)=x(450-2x)](https://img.qammunity.org/2019/formulas/mathematics/college/94syjg0cyji60ctyloj8kyrad10v9jbxcn.png)
. The easiest way is to use the formula for the vertex,
![x= (-b)/(2a) = 450/4 = 112.5 \ m](https://img.qammunity.org/2019/formulas/mathematics/college/ldq5tn9c4uyadbhzsq05hbuuv7pzdgw7al.png)
, and
![f( (-b)/(2a))=f(112.5)=25,312.5 \ m^2](https://img.qammunity.org/2019/formulas/mathematics/college/wn6qaixbt6q8zkw518pw2ia3bzwuhx59rq.png)
.
I realize it sounds like a big number, but the largest area that can be enclosed is 25,312.5 m^2 (if I did this correctly!).