x, y - the lengths of the sides of the lot
The area of a rectangle is the product of the lengths of its sides. The area is 240 m².
[tex]xy=240[/tex]
It requires 64 m of fencing materials to enclose it, so its perimeter is 64 m. The perimeter of a rectangle is two times the sum of the lengths of its sides.
[tex]2(x+y)=64 \ \ \ \ \ \ |\div 2 \\
x+y=32 \ \ \ \ \ \ \ \ \ \ |-x \\
y=32-x[/tex]
Substitute 32-x for y in the first equation:
[tex]x(32-x)=240 \\
32x-x^2=240 \\
-x^2+32x-240=0 \\
-x^2+12x+20x-240=0 \\
-x(x-12)+20(x-12)=0 \\
(-x+20)(x-12)=0 \\
-x+20=0 \ \lor \ x-12=0 \\
x=20 \ \lor \ x=12
[/tex]
[tex]y=32-x \\
\hbox{for } x=20: \\
y=32-20=12 \\ \\
\hbox{for } x=12: \\
y=32-12=20[/tex]
The dimensions of the lot are 12 m by 20 m.
The width of the lot is 12 m.
