Let:
[tex]\begin{gathered} w=width \\ l=\text{length}=60-2w \\ \text{ Since it is a rectangle, the area is:} \\ A=w(60-2w)=60w-2w^2 \\ \end{gathered}[/tex]Let's express the equation as:
[tex]A(w)=-2w^2+60w[/tex]We can find the vertex of this parabola (peak, or value where the area is maximum) using the following formula:
[tex]\begin{gathered} x_{\max }=\frac{-b}{2(a)} \\ \text{where:} \\ b=60 \\ a=2 \\ x_{\max }=\frac{-60}{-4}=15 \end{gathered}[/tex]And the maximum area will be:
[tex]A(15)=-2(15)^2+60(15)=-450+900=450m^2[/tex][tex]\begin{gathered} w=15 \\ \text{ Since:} \\ l=60-2w \\ l=60-30=30 \\ \text{width}=15 \\ \text{length}=30 \end{gathered}[/tex]