Answer
$1000$ feet by $1500$ feet
Work Step by Step
Let $x$ and $y$ be the field's dimensions.
$xy$ = $1.5{\times}{10^{6}}$
$y$ = $\frac{1.5{\times}{10^{6}}}{x}$
We have to minimize the amount of fencing, which is
$F(x)=3x+2y$ = $3x+2(\frac{1.5{\times}{10^{6}}}{x})$ = $3x+\frac{3{\times}{10^{6}}}{x}$
$F'(x)$ = $3-\frac{3{\times}{10^{6}}}{x^{2}}$ = $\frac{3(x^{2}-10^{6})}{x^{2}}$
The critical number is $x$ = $10^{3}$ and
$F'(x)$ $\lt$ $0$ for $0$ $\lt$ $x$ $\lt$ $10^{3}$x
$F'(x)$ $\gt$ $0$ if $x$ $\gt$ $10^{3}$
so the absolute minimum occurs when $x$ = $10^{3}$ and $y$ = $1.5{\times}{10^{3}}$
The field should be $1000$ feet by $1500$ feet with the middle fence parallel to the short side of the field
