Now that I understand your "layout", yes, the formulae you created will work just fine, remembering that little "y" is the dimension of the length of a smaller rectangle, and $Y = 3y$ the dimension of the length of the large rectangle encompassing the whole.
$$ 4x+3y = 1080,~~~\text{Area} = 3xy.$$
We can then express Area as a function of $x$, first solving for $y$ in the first equation, and substituting this expression for $y$ into the Area equation:
$$4x + 3y = 1080 \iff 3y = 1080 - 4x \iff y = 360 - \frac 43 x$$
$$\text{Area}\;= 3x(360 - (4/3)x) = x(1080 - 4x) = = 1080x - 4x^2$$
To maximize area, we calculate $A'$ and set $A' = 0$ to determine the function's critical point:
$$A' = 1080 - 8x,\qquad A' = 0 \implies 1080 - 8x = 0 \implies x = 135. $$ We can easily show that the solution to $A' = 0 \iff x = 135$ ft. does in fact give the maximum value for $A$.
Now we have $x$, and will need only to compute $y = 360 - 4/3(135) = 360 - 180 = 180$ ft.
Then $Y = 3 \times 180 = 540$ ft, and so we've determined the larger rectangle's dimensions for maximized area: $ Y$ ft $ \times x$ ft $= 540 \text{ ft} \times 135 \text{ ft}$, with smaller rectangles each $180 \text{ ft} \times 135\text{ ft}$
Your first result is correct. It is given (or at least strongly implied) that the sides of the rectangle are growing at a constant rate. Your second attempt is assuming that the area is also growing at a constant rate, but that isn't true. To check, you can calculate how fast the area of the rectangle is growing at 2 seconds by feeding that new info into the equation you came up with in the first part.
Best Answer
Hint. Because you have to use the chain rule (then the product rule). Both $w$ and $l$ are functions of time. So when you differentiate with respect to $t$, you have to apply the chain rule as it is a function of time. Do the same with $l$. Just like taking the derivate of $h(x) g(x)$.