Question

A farmer has 1800 feet of fence and wants to enclose a rectangular field with two internal fences parallel to one side. What are the dimensions that would maximize the area?

Answer 1

So.. if you notice the picture below.. .the length is that much

thus    $\bf A=lw\qquad \cfrac{1800-3w}{2}=l\implies 900-\cfrac{3w}{2}=l \\\\\\ A=\left( 900-\cfrac{3w}{2} \right)w\implies A=900w-\cfrac{3w^2}{2}\implies A=900w-\cfrac{3}{2}w^2$

take the derivative of A, zero it out, and then check the critica points for any maxima by using the first-derivative test