Word problem finding dimensions (please check my answer)

[pita0001]

A homeowner wants to fence a rectangular play yard using 72 ft of fencing. The side of the house will be used as one side of the rectangle. Find the dimensions for which the area is a maximum and determine the maximum area.
I got L=18 and W=36 So my maximum area is 648 ftIs this correct?

 

[MarkFL]

Well, let's see. Suppose the total length of fencing available is $L$. Let's let $y$ be the length of the two sides perpendicular to the house and $x$ be the length of the side parallel to the house. So we have the constraint:

$$x+2y=L$$

And the objective function, that which we wish to maximize is, which is the area $A$ of the enclosed area:

$$A=xy$$

Solving the constraint for $x$, we obtain:

$$x=L-2y$$

And so substituting for $x$ into the objective function, we get:

$$A=(L-2y)y$$

Now, we see that this function is quadratic, and has the roots:

$$y=0,\,\frac{L}{2}$$

We know this quadratic function opens downward, and so the vertex is at the maximum, and this vertex will lie on the axis of symmetry, which is midway between these roots, and so the function is maximized for:

$$y=\frac{L}{4}\implies x=\frac{L}{2}$$

And the maximum value of the objective function is therefore:

$$A_{\max}=\frac{L^2}{8}$$

Now, using the given value of $$L=72\text{ ft}$$, we then have:

$$x=\frac{72\text{ ft}}{2}=36\text{ ft}$$

$$y=\frac{72\text{ ft}}{4}=18\text{ ft}$$

$$A_{\max}=\frac{(72\text{ ft})^2}{8}=648\text{ ft}^2$$

So, yes you are correct.