To answer this question, we first need to determine the objective and constraint equations.
The objective is to maximize the area of the rectangular garden, which is given by the equation $A = xy$.
The constraint is that the total cost of the fencing must not exceed $320. In other words, we have the equation $6x + 2(y+x) \le 320$.
To express the quantity to be maximized as a function of $x$, we can simply plug in the equation for the area of the rectangular garden into the objective equation. This gives us the function $f(x) = xy = x^2y$.
To find the optimal values of $x$ and $y$, we can use the method of Lagrange multipliers. Let $\lambda$ be the Lagrange multiplier. The Lagrange function is given by
$$L(x,y,\lambda) = x^2y + \lambda(6x + 2(y+x) - 320)$$
To find the optimal values of $x$ and $y$, we need to find the values of $x$ and $y$ that maximize $L(x,y,\lambda)$. To do this, we take the partial derivatives of $L(x,y,\lambda)$ with respect to $x$ and $y$ and set them equal to 0. This gives us the equations
$$\frac{\partial L}{\partial x} = 2xy + 6\lambda = 0 \quad \text{and} \quad \frac{\partial L}{\partial y} = x^2 + 2\lambda = 0$$
We can solve these equations simultaneously to find the optimal values of $x$ and $y$. First, we can solve for $y$ in the second equation to get
$$y = -\frac{x^2}{2\lambda}$$
Substituting this expression for $y$ into the first equation, we get
$$2x \left(-\frac{x^2}{2\lambda}\right) + 6\lambda = 0 \quad \Rightarrow \quad -x^3 + 12\lambda^2 = 0$$
Since $\lambda$ cannot be 0, we can divide both sides of this equation by $-12\lambda^2$ to get
$$x^3 = 12\lambda^2 \quad \Rightarrow \quad x = \sqrt[3]{12\lambda^2}$$
Substituting this expression for $x$ into the expression for $y$, we get
$$y = -\frac{\sqrt[3]{12\lambda^2}}{2\lambda}$$
Now we need to find the value of $\lambda$ that satisfies the constraint $6x + 2(y+x) \le 320$. Substituting the expressions for $x$ and $y$ into this equation, we get
$$6\sqrt[3]{12\lambda^2} + 2\left(-\frac{\sqrt[3]{12\lambda^2}}{2\lambda} + \sqrt[3]{12\lambda^2}\right) \le 320$$
Simplifying, we get
$$-\frac{\sqrt[3]{12\lambda^2}}{2\lambda} + 10\sqrt[3]{12\lambda^2} \le 320$$
Dividing both sides by $10\sqrt[3]{12\lambda^