The question is as follows: Find the area of the largest rectangle that has sides parallel to the coordinate axes, one corner at the origin and the opposite corner on the line 3x+2y=12 in the first quadrant.
I get that the equation I have to maximize is in the form of A=bh but I don't know how to eliminate one of the variables to continue.
On
Any point on the line can be represented as (x,(12-3x)/2)
Now we have to find the maximum area of a rectangle given one point as (0,0) and opposite point as $(x,(12-3x)/2$)
Area=$(x)*(12-3x)/2$
We have to maximize this area taking derivative and equating to $0$ we get value of $x=2$
Substituting this value in area we get area=6 which should be maximum
On
Suppose that your rectangle has vertices $(0, 0)$, $(x, 0)$, $(0, y)$, and $(x, y)$, where $x > 0$, $y > 0$, and $$ 3x + 2y = 12. \tag{1} $$ Then the area of your rectangle is given by $$ A = xy. \tag{2} $$ But (1) implies that $$ y = 6 - \frac{3x}{2}. \tag{3} $$ Putting the value of $y$ from (3) into the formula in (2), we obtain $$ A = A(x) = x \left( 6 - \frac{3x}{2} \right) = 6x - \frac{3x^2}{2}. \tag{4}$$ Now (4) gives area $A$ as a function of $x$ for $x > 0$.
Differentiating both sides of (4) w.r.t. $x$ we obtain $$ A^\prime(x) = 6 - 3x. $$ Thus we see that $$ A^\prime(x) \ \begin{cases} > 0 \ & \ \mbox{ for } x < 2, \\ = 0 \ & \ \mbox{ for } x = 2, \\ < 0 \ & \ \mbox{ for } x > 2. \end{cases} $$ Thus the area attains its (relative) maximum value at $x = 2$, and since this is the only relative extreme value of $A$, this is in fact the absolute maximum value of $A$.
Therefore the largest possible area is given by $$ A(2) = 12 - 6 = 6. $$
Hope this solves your problem.
Since the bottom left corner of the rectangle is at the origin, then the $(x,\,y)$ coordinates of the top right corner will be the base and height (draw a figure to help visualize). We know that this point is on the line $3x+2y=12$, so that $y=-\frac{3}{2}x+6$. Plugging this in gives $A=x(-\frac{3}{2}x+6)$, which has only one variable.