A rectangular lot has a river on one side and is fenced off into 3 areas with 752m of fencing. Find the dimensions and its maximum area. Note: There is no fencing along the river

I have:

Perimeter = 752m
$\displaystyle 3w + 2l = 752$
$\displaystyle 2l = 752 - 3w$
$\displaystyle l = 376 - 3/2w$

Area = l x w
$\displaystyle A = (376-3/2w)(w)$
$\displaystyle A = [376w-(3/2w^2)]$

Thats what i have so far, don't know if its right or wrong. Would be gratefull if anyone could tell me if what i have so far looks right, and what/how to do next.

2. Very good so far.
Except that 3/2w is misleading. (3/2)w is better.
So is (3/2)w^2.

A = 376w -(3/2)w^2
You transform that into
A = a(w-b) +c
so that you can get the vertex (b,c),
where b is the value of w at the vertex.
and c is the value of A at the vertex.

A = 376w -(3/2)w^2
A = -(3/2)[w^2 -(376w)/(3/2)]
A = -(3/2)[w^2 -(376*2 /3)w +(376/3)^2] -(3/2)(-(376/3)^2)

. Why don't we use decimal points?

A = -(1.5)[(w -125.333)^2] +23,562.667 ----------(i)

Parabola (i) is vertical, that opens downward because of the negative coefficient of w^2.
Therefore the vertex is the highest point, a maximum A.

Hence, for maximum A,