Perimeter of fencing $=2400$

Area $=?$

$\mathrm{A}= base \times height =\mathrm{xy}$

We know that the perimeter of fence $=2400$. In our case that means $2 \mathrm{x}+\mathrm{y}=2400$. This tells us $y=2400-2 x$

Therefore area can be written as $\mathrm{A}=\mathrm{x}(2400-2 \mathrm{x})=2400 \mathrm{x}-2 \mathrm{x}^{2}$

$A^{\prime}=2400-4 x$

$2400-4 x=0$

$x=600$

$\mathrm{A}^{\prime}(\mathrm{x})=2400-4 \mathrm{x}$

$\mathrm{A}^{\prime}(500)=2400-4(500)=400$

$\mathrm{~A}^{\prime}(700)=2400-4(700)=-400$

If width $(\mathrm{x})=600$ feet,

then length $(\mathrm{y})=2400-2 \mathrm{x}=2400-1200=1200$ feet

Thus the rectangular field should be $\mathbf{6 0 0}$ feet wide and $\mathbf{1 2 0 0}$ feet long.