*Appendix A.1*

In this section, we provide the details to the derivation of solution to the utility maximization problem for semi-specialized mode (*xy*/*y*)1, as presented in Section 3.2. Note that the case (*xy*/*y*)1 corresponds to *Zi* = (*xi*, *xsi* , *xdi* , *yi*, *ysi* , *ydi* ) = (+, 0, +, +, +, <sup>0</sup>). Hence, the constrained utilities maximization problem can be formulated as

$$\max\_{x\_1, y\_1, x\_1^s, y\_1^d, I\_{1x}, I\_{1y}} lI\_1 = x\_1^\beta (y\_1 + ky\_1^d)^{1-\beta} \tag{A1}$$

$$\text{s.t.} \quad \mathbf{x}\_1 + \mathbf{x}\_1^s = a\_{1x} l\_{1x}, \quad y\_1 = a\_{1y} l\_{1y}, \quad l\_{1x} + l\_{1y} = 1, \quad y\_1^d = p \mathbf{x}\_1^s. \tag{A2}$$

Note that the four constraints allow us to express the variables *x*1, *y*1, *<sup>l</sup>*1*y*, and *yd*1 in terms of *xs*1 and *l*1*x*. Then, the constraint optimization problem in Equations (A1) and (A2) can be rewritten as

$$\max\_{x\_{1^s}^s, l\_{1x}} \mathcal{U}\_1 = (a\_{1x}l\_{1x} - x\_1^s)^\beta (a\_{1y}(1 - l\_{1x}) + kpx\_1^s)^{1-\beta} \tag{A3}$$

$$0 \le \mathbf{t}. \quad 0 \le x\_1^s \le a\_{1x} l\_{1x} \quad \text{and} \quad 0 \le l\_{1x} \le 1. \tag{A4}$$

First-order derivatives on the utility function *U*1 gives us

$$\frac{\partial \mathcal{U}\_1}{\partial \mathbf{x}\_1^s} = \left( -\frac{\beta}{a\_{1x}l\_{1x} - \mathbf{x}\_1^s} + \frac{kp(1-\beta)}{a\_{1y}(1-l\_{1x}) + kpx\_1^s} \right) \mathcal{U}\_1,\tag{A5}$$

$$\frac{\partial \mathcal{U}\_{1}}{\partial l\_{1x}} = \left(\frac{a\_{1x}\beta}{a\_{1x}l\_{1x} - \mathbf{x}\_{1}^{s}} - \frac{a\_{1y}(1 - \beta)}{a\_{1y}(1 - l\_{1x}) + kpx\_{1}^{s}}\right) \mathcal{U}\_{1}.\tag{A6}$$

*Mathematics* **2019**, *7*, 1152

Setting both derivatives equal to zero yields the condition on relative price *p* = *<sup>a</sup>*1*y*/(*ka*1*x*). With this relative price, Equation (A5) then simplifies to *xs*1 = *<sup>a</sup>*1*x*(*l*1*x* − *β*). Moreover, the constraints in Equation (A4) are equivalent to *β* ≤ *l*1*x* ≤ 1. Furthermore, the original constraints in Equation (A2) lead to

$$\mathbf{x}\_{1} = a\_{1x}l\_{1x} - \mathbf{x}\_{1}^{s} = \beta a\_{1x} \tag{A7}$$

$$y\_1 = a\_{1y} l\_{1y} = a\_{1y} (1 - l\_{1x}) \tag{A8}$$

$$y\_1^d = p \\ x\_1^s = \frac{a\_{1y}(l\_{1x} - \beta)}{k}.\tag{A9}$$

Substituting the appropriate variables in the utility function leads to

$$
\beta L\_1 = \beta^{\beta} (1 - \beta)^{1 - \beta} a\_{1x}^{\beta} a\_{1y}^{1 - \beta} \tag{A10}
$$

*Appendix A.2*

In this section, we explain the solution derivation to the utility maximization problem for fully specialized mode (*x*/*y*)1, in the context of Structure C, i.e.,

$$\max\_{x\_1, x\_1^s, y\_1^d} \mathcal{U}\_1 = x\_1^\beta (k y\_1^d)^{1-\beta},\tag{A11}$$

$$\text{s.t.} \quad \mathbf{x}\_1 + \mathbf{x}\_1^s = a\_{1x} l\_{1x}, \quad y\_1^d = p \mathbf{x}\_1^s, \quad l\_{1x} = 1. \tag{A12}$$

Replacing *x*1 and *yd*1 in terms of *<sup>x</sup>s*1, the constrained optimization problem in Equations (A11) and (A12) may be reformulated as

$$\max\_{0 \le x\_1^s \le a\_{1x}} lI\_1 = (a\_{1x} - x\_1^s)^\beta (kpx\_1^s)^{1-\beta}. \tag{A13}$$

The first-order derivative of the utility function reads

$$\frac{\mathrm{d}\mathrm{d}I\_{1}}{\mathrm{d}\mathbf{x}\_{1}^{s}} = \left(-\frac{\beta}{a\_{1\mathbf{x}} - \mathbf{x}\_{1}^{s}} + \frac{1-\beta}{\mathbf{x}\_{1}^{s}}\right) \mathrm{d}I\_{1}.\tag{A14}$$

At the critical point, the first-order derivative vanishes, and this implies

$$
\alpha\_1^s = (1 - \beta) a\_{1x} \,\tag{A15}
$$

Furthermore, we have

$$x\_1 = \beta a\_{1x} \quad \text{and} \quad y\_1^d = (1 - \beta) p a\_{1x}. \tag{A16}$$

The value of the utility function is

$$
\Delta l\_1 = \beta^{\beta} (1 - \beta)^{1 - \beta} (pk)^{1 - \beta} a\_{1\mathbf{x}}.\tag{A17}
$$

In the context of Structure C, we have the market clearing condition *xs*1 = *xd*2, which sets the equilibrium relative price *p* = *β<sup>a</sup>*2*y* (<sup>1</sup>−*β*)*<sup>a</sup>*1*x* . Hence, the critical utility value for Farmer 1 in Structure C is *Uc*1 = *<sup>β</sup>aβ*1*x*(*ka*2*y*)<sup>1</sup>−*β*. The corresponding utility of Farmer 2 is *Uc*2 = (1 − *β*)(*ka*1*x*)*βa*1−*<sup>β</sup>* 2*y* .
