Farmers’ adoption of organic production

Abstract

This paper presents a theoretical model of the farmer’s decision to adopt alternative technologies in agriculture. The decision concerns the allocation of lands for old and new technology. We consider the case where organic production requires adoption of a new/alternative technology to be profitable. This paper suggests that an entirely theoretical exercise can illuminate parts of this complex issue which empirical work has not been able to reach. We show the importance of (1) the available quantity of land devoted to agriculture, (2) the productivity of the new/alternative technology, (3) the incentive mechanism and finally, (4) the constraints on output of the new/alternative technology. The results of this article are not confined to the agricultural sector. The insights apply to other technology decisions.

Appendix

1.1 Appendix 1: Proof of Lemma 2.5

Proof

(i) is obvious.

(ii) Take $x_0 >0$. We have for $x>0$, $\Phi (x) \leqslant \Phi (x_0) +\Phi ^\prime (x_0) (x-x_0)$ since $\Phi$ is concave. Hence ${{x^{{1}\over {\alpha }}}\over {\Phi (x)}}\geqslant {{x^{{1}\over {\alpha }}}\over {\Phi (x_0) +\Phi ^\prime (x_0) (x-x_0)}} \rightarrow +\infty$ when $x\rightarrow +\infty$.

We have ${{R}\over {\Phi (A)}}= P_2 ^{\frac{1}{\alpha }} \alpha \left\{ \frac{1-\alpha }{w}\right\} ^{\frac{(1-\alpha )}{\alpha }}{{A^{{1}\over {\alpha }}}\over {\Phi (A)}} - \; P_1^{\frac{1}{\alpha }} \alpha \left\{ \frac{1-\alpha }{w}\right\} ^{\frac{(1-\alpha )}{\alpha }} {{1}\over {\Phi (A)}}$.

The first term of the RHS converges to $+\infty$ while the second term converges to a finite value since ${{1}\over {\Phi (A)}}\rightarrow {{1}\over {\Phi (+\infty )}}<+\infty$, the conclusion follows. $\square$

1.2 Appendix 2: Proof of Proposition 2.1

Proof

Result (i) is easy to prove.

To prove result (ii), since the land constraint is binding, profit becomes

$$\begin{aligned} \pi = \{P_1 S_1^\alpha L_1^{1-\alpha } - wL_1 - C(S_1) + P_2 A {{(V-S_1)}}^\alpha L_2^{1-\alpha } -wL_2 - \Phi (A) C(V-S_1)\}. \end{aligned}$$

• Consider the case $\gamma \Phi (A) V-R > 0$ and $\gamma V+ R>0$. We claim that in this case the solution is interior. Suppose our claim is true. The FOC are as follows:

  $$\begin{aligned}\frac{\partial \pi }{\partial S_1}&= P_1 \alpha S_1^{\alpha -1} L_1^{1-\alpha } - C^\prime (S_1) - P_2 A \alpha (V-S_1)^{\alpha -1} L_2^{1-\alpha } \\ &\quad +\, \Phi (A) C^\prime (V-S_1) =0 \end{aligned}$$

  (5)
  $$\begin{aligned}\frac{\partial \pi }{\partial L_1}= P_1 (1-\alpha )S_1^\alpha L_1^{-\alpha } - w=0 \end{aligned}$$

  (6)

  and

  $$\begin{aligned} \frac{\partial \pi }{\partial L_2}={{ P_2 A }}(1-\alpha )(V-S_1)^\alpha L_2^{-\alpha } - w=0. \end{aligned}$$

  (7)

  From (6) and (7), it can be written as

  $$\begin{aligned}\left( \frac{L_1}{S_1} \right) ^{1-\alpha } = \left[ \frac{P_1 (1-\alpha )}{w}\right] ^{\frac{1-\alpha }{\alpha }}, \end{aligned}$$

  (8)
  $$\begin{aligned}\left( \frac{L_2}{V-S_1} \right) ^{1-\alpha } = \left[ \frac{P_2 A (1-\alpha )}{w}\right] ^{\frac{1-\alpha }{\alpha }}. \end{aligned}$$

  (9)

  Equation (5) is also written as

  $$\begin{aligned} P_1 \alpha \left( \frac{L_1}{S_1} \right) ^{1-\alpha } -P_2 A \alpha \left( \frac{L_2}{V-S_1} \right) ^{1-\alpha } = \gamma S_1 - \gamma \Phi (A) (V-S_1) \end{aligned}$$

  (10)

  where we use $C(S)= \gamma \frac{S^2}{2}$. Substituting (8), (9) in to (10) yields

  $$\begin{aligned} \alpha \left\{ \frac{1-\alpha }{w}\right\} ^{\frac{(1-\alpha )}{\alpha }} \left[ P_1 ^{\frac{1}{\alpha }}- (P_2 A)^{\frac{1}{\alpha }}\right] +\gamma \Phi (A) V = \gamma [1+\Phi (A) ]S_1. \end{aligned}$$

  Recall that

  $$\begin{aligned}R= -\alpha \left\{ \frac{1-\alpha }{w}\right\} ^{\frac{(1-\alpha )}{\alpha }} \left[ P_1 ^{\frac{1}{\alpha }}- (P_2 A)^{\frac{1}{\alpha }}\right] . \end{aligned}$$

  Then the FOC for the optimal $S_1$ is $\gamma \Phi (A) V-R=\gamma [1+\Phi (A) ]S_1.$ We then get

  $$\begin{aligned} S^{*}_{1} = \frac{\gamma \Phi (A)V - R}{\gamma (1+\Phi (A))}. \end{aligned}$$

  (11)

  The binding land constraint implies

  $$\begin{aligned} S^{*}_{2} =V-S^* _1= \frac{\gamma V + R}{\gamma (1+\Phi (A))}. \end{aligned}$$

  (12)

  From (8), (9) we obtain

  $$\begin{aligned} L^{*}_{1} = \left[ \frac{\gamma \Phi (A)V - R}{\gamma (1+\Phi (A))}\right] \left\{ \frac{P_1(1-\alpha )}{w}\right\} ^{\frac{1}{\alpha }} \end{aligned}$$

  (13)

  and

  $$\begin{aligned} L^{*}_{2} = \left[ \frac{\gamma V + R}{\gamma (1+\Phi (A))}\right] \left\{ \frac{P_2A(1-\alpha )}{w}\right\} ^{\frac{1}{\alpha }}. \end{aligned}$$

  (14)

  We have proved that our claim is true.

• Consider the case $\gamma \Phi (A) V-R\leqslant 0$. We claim that $S^* _1 =0$. From the proof given above $S^* _1 =0$ or $S^* _1=V$. Let $\pi ^0$ and $\pi ^V$ denote respectively the profits corresponding to $S^* _1= 0$ and $S^* _1= V$. In the first case we have $S^* _2 = V$, $L^* _1 = 0, L^* _2 = V \left[ \frac{P_2 A (1-\alpha )}{w} \right] ^{\frac{1}{\alpha }}$ while in the second case $S^* _2 = 0$, $L^*_ 2 = 0, L^*_1 = V \left[ \frac{P_1 (1-\alpha )}{w} \right] ^{\frac{1}{\alpha }}$. Recall that $R= \alpha \left\{ \frac{1-\alpha }{w}\right\} ^{\frac{(1-\alpha )}{\alpha }} \left[ (P_2 A)^{\frac{1}{\alpha } }- P_1 ^{\frac{1}{\alpha } } \right]$.

  Let $\Delta = \pi ^0 -\pi ^V$. Tedious computations give

  $$\begin{aligned} \Delta=\, & {} P_2 A V^\alpha \left[ \frac{P_2 A (1-\alpha )}{w} \right] ^{\frac{1-\alpha }{\alpha }} V^{1-\alpha } - w \left[ \frac{P_2 A (1-\alpha )}{w} \right] ^{\frac{1}{\alpha } }V -\frac{\Phi (A) \gamma V^2}{2} \\ &-\, P_1 V^\alpha \left[ \frac{P_1 (1-\alpha )}{w} \right] ^{\frac{1-\alpha }{\alpha }} V^{1-\alpha } + w \left[ \frac{P_1 (1-\alpha )}{w} \right] ^{\frac{1}{\alpha } }V +\frac{ \gamma V^2}{2}\\=\, & {} V w {\frac{\alpha -1}{\alpha }} (1-\alpha ) ^{\frac{1 }{\alpha }} (\frac{\alpha }{1-\alpha } )\left[ (P_2 A)^{\frac{1 }{\alpha }} -P_1 ^{\frac{1 }{\alpha }}\right] + \frac{V}{2} (V\gamma +R)\\ &-\,\frac{V}{2}(\gamma \Phi (A)V-R)\\=\, & {} V \alpha \left\{ \frac{1-\alpha }{w}\right\} ^{\frac{(1-\alpha )}{\alpha }} \left[ (P_2 A)^{\frac{1 }{\alpha }} -P_1 ^{\frac{1 }{\alpha }}\right] + \frac{V}{2} (V \gamma - \Phi (A) \gamma V) \\=\, & {} {{RV+ \frac{V}{2}(V \gamma - \Phi (A)\gamma V) }}\\= & {} {{\frac{V}{2} (V \gamma + R)- \frac{V}{2}(\Phi (A) \gamma V-R)}}. \end{aligned}$$

  If $\gamma \phi (A)V-R\leqslant 0$ then $R>0$ and $V\gamma +R >0$. This implies $\Delta >0$. Hence $S^* _1=0$ which implies

  $$\begin{aligned} S^{*}_{2}=V,\quad L^{*}_{1}=0,\quad L^{*}_{2} = V \left\{ \frac{P_2 A (1-\alpha )}{w}\right\} ^{\frac{1}{\alpha }}. \end{aligned}$$

• Consider the case $\gamma V+ R\leqslant 0$. We then have $R<0$. This implies $\Delta <0$. Hence $S^{*}_1=V$ and

  $$\begin{aligned} S^{*}_{2}=0,\quad L^{*}_{2 }=0,\quad L^{*}_{1} = V \left\{ \frac{P_1 (1-\alpha )}{w}\right\} ^{\frac{1}{\alpha }}. \end{aligned}$$

$\square$

1.3 Appendix 3: Proof of Proposition 2.2

Proof

When $\frac{R}{\gamma \Phi (A) }< V < Q(P_1, P_2, \alpha , w, \Phi , \gamma , A)$ or $-\frac{R}{\gamma }<V < Q(P_1, P_2, \alpha , w, \Phi , \gamma , A)$ from Proposition 2.1, we have

$$\begin{aligned} S^{*}_{2} = \frac{\gamma V + R}{\gamma (1+\Phi (A))},\quad R= \alpha \left\{ \frac{1-\alpha }{w}\right\} ^{\frac{(1-\alpha )}{\alpha }} \left[ (P_2 A)^{\frac{1}{\alpha } }- P_1 ^{\frac{1}{\alpha } } \right] . \end{aligned}$$

We can write $S^{*}_{2} = \frac{V(1 + \frac{R}{\gamma V})}{1+\Phi (A)}$.

Consider first the case $\frac{R}{\gamma \Phi (A) }< V < Q(P_1, P_2, \alpha , w, \Phi , \gamma , A)$.

We observe that $\frac{R}{\gamma V} - \Phi (A) < 0$. We obtain

$$\begin{aligned} \text{Log} S^{*}_{2}=\, & {} \text{Log} V + \text{Log} \left( 1+\frac{R}{\gamma V}\right) - \text{Log} (1+\Phi (A))\\ \frac{\mathrm{d}}{{\mathrm{d}}A} \text{Log} S^{*}_{2}=\, & {} \frac{1}{(1 + \frac{R}{\gamma V})} \frac{1}{\gamma V}\left[ P_2^{{1}\over {\alpha }} \left( \frac{1-\alpha }{w}\right) ^{\frac{(1-\alpha )}{\alpha }} A^{\frac{(1-\alpha )}{\alpha }}\right] - \frac{1}{1+\Phi (A)}\Phi '(A)\\=\, & {} \frac{1}{\gamma V + R} P_2^{{1}\over {\alpha }} \left[ \frac{1-\alpha }{w}\right] ^{\frac{(1-\alpha )}{\alpha }} A^{\frac{(1-\alpha )}{\alpha }} - \frac{\Phi '(A)}{1+\Phi (A)}. \end{aligned}$$

Observe that $\frac{1}{\gamma V + R}>\frac{1}{\gamma V(1+\Phi (A))}$. Hence,

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}A} \text{Log} S^{*}_{2}>\frac{1}{1+\Phi (A)}\left[ \frac{1}{\gamma V} P_2^{{1}\over {\alpha }} \left( \frac{1-\alpha }{w}\right) ^{\frac{(1-\alpha )}{\alpha }} A^{\frac{(1-\alpha )}{\alpha }} - {\Phi '(A)}\right] . \end{aligned}$$

Let

$$\begin{aligned} \phi (A) = \frac{1}{\gamma V} P_2^{{1}\over {\alpha }} \left[ \frac{1-\alpha }{w}\right] ^{\frac{(1-\alpha )}{\alpha }} A^{\frac{(1-\alpha )}{\alpha }} - {\Phi '(A)}. \end{aligned}$$

The function $\phi$ is increasing. It takes a negative value $-\Phi '(0)$ when $A=0$ and equals $+\infty$ when $A=+\infty$. Hence, there exists a value $\hat{A}$ such that if $A>\hat{A}$ then $\phi (A)>0$ implying $\frac{{\mathrm{d}}}{{\mathrm{d}}A} \text{Log} S^{*}_{2}>0$.

When $A\rightarrow 0$, $\frac{1}{\gamma V + R} P_2^{{1}\over {\alpha }} [\frac{1-\alpha }{w}]^{\frac{(1-\alpha )}{\alpha }} A^{\frac{(1-\alpha )}{\alpha }} - \frac{\Phi '(A)}{1+\Phi (A)}\rightarrow -\Phi '(0)<0.$ Hence, there exists $\tilde{A}$ such that, if $A<\tilde{A}$ then $\frac{{\mathrm{d}}}{{\mathrm{d}}A} \text{Log} S^{*}_{2}<0$. The proof is similar for the case $-\frac{R}{\gamma }<V < Q(P_1, P_2, \alpha , w, \Phi , \gamma , A)$. $\square$

1.4 Appendix 4: Proof of Lemma 2.6

Proof

• (i) is obvious.

• (ii.a) We have $Q(P_1, P_2, \alpha , w, \Phi , \gamma , A) = V_1 \left[ 1+ {{1}\over {\Phi (A)}}\left( {{A}\over {A_1}}\right) ^{{1}\over {\alpha }}\right] .$

  Hence,

  $$\begin{aligned} V<Q(P_1, P_2, \alpha , w, \Phi , \gamma , A)\Leftrightarrow & {} \left[ {{V}\over {V_1}} -1\right] < {{1}\over {A_1}^{{1}\over {\alpha }}} {{A^{{1}\over {\alpha }}}\over {\Phi (A)}}\\\Leftrightarrow & {} {{A^{{1}\over {\alpha }}}\over {\Phi (A)}} > {A_1}^{{1}\over {\alpha }} \left[ {{V}\over {V_1}} - 1\right] . \end{aligned}$$

  Let $\phi (A)= {{A^{{1}\over {\alpha }}}\over {\Phi (A)}}.$ Since $\Phi$ is concave, the function $\phi$ is increasing. Let $A_2$ be defined by

  $$\begin{aligned} {{{A_2}^{{1}\over {\alpha }}}\over {\Phi (A_2)}}= A_1^{{1}\over {\alpha }} \left[ {{V}\over {V_1}}-1 \right] . \end{aligned}$$

  Then obviously, $V\leqslant Q(P_1, P_2, \alpha , w, \Phi , \gamma , A) \Leftrightarrow A \geqslant A_2.$

  If ${{A_1^{{1}\over {\alpha }}}\over {\Phi (A_1)}} < A_1^{{1}\over {\alpha }} \left[ {{V}\over {V_1}}-1 \right]$ or equivalently ${{1}\over {\Phi (A_1)}}< \left[ {{V}\over {V_1}} -1\right]$ then $A_2 >A_1$ since function $\phi$ is increasing. If ${{1}\over {\Phi (A_1)}}= \left[ {{V}\over {V_1}} -1\right]$ then $A_2 =A_1$ and obviously ${{1}\over {\Phi (A_1)}}> \left[ {{V}\over {V_1}} -1\right]$ then $A_2 <A_1.$

• (ii.b) Let $\psi (A)= {{A^{{1}\over {\alpha }}-A_1^{{1}\over {\alpha }}}\over {\Phi (A)}} {{1}\over {A_1 ^{{1}\over {\alpha }}}}.$ Function $\psi$ is increasing since $\phi$ is increasing. It satisfies $\psi (0)= -\infty ,\; \psi (A_1)=0$. Therefore, $A_3 >A_1$ since $\psi (A_3)= {{V}\over {V_1}}$. Now, observe that

  $$\begin{aligned} {{A_3^{{1}\over {\alpha }}}\over {\Phi (A_3)}}= A_1^{{1}\over {\alpha }} {{V}\over {V_1}}+ {{A_1^{{1}\over {\alpha }}}\over {\Phi (A_3)}} \end{aligned}$$

  while $A_2$ verifies ${{A_2^{{1}\over {\alpha }}}\over {\Phi (A_2)}}= A_1^{{1}\over {\alpha }} \left[ {{V}\over {V_1}}-1 \right] ,$ hence $A_3 >A_2$.

  Since $R= \alpha \left\{ \frac{1-\alpha }{w}\right\} ^{\frac{(1-\alpha )}{\alpha }} \left[ (P_2 A)^{\frac{1}{\alpha } }- P_1 ^{\frac{1}{\alpha } } \right] ,$ we get ${{R}\over {\gamma \Phi (A)}}= {{1}\over {A_1 ^{{1}\over {\alpha }}}} \left[ {{A ^{{1}\over {\alpha }}-A_1 ^{{1}\over {\alpha }}}\over {\Phi (A)}} \right] V_1.$

  We have

  $$\begin{aligned} {{{{R}\over {\gamma \Phi (A)}} \leqslant V \Leftrightarrow \psi (A) \leqslant {{V}\over {V_1}}=\psi (A_3) \Leftrightarrow A \leqslant A_3 }} \end{aligned}$$

• (ii.c) Since $R= \alpha \left\{ \frac{1-\alpha }{w}\right\} ^{\frac{(1-\alpha )}{\alpha }} \left[ (P_2 A)^{\frac{1}{\alpha } }- P_1 ^{\frac{1}{\alpha } } \right] ,$ we have $-{{R}\over {\gamma }} = V_1 \left[ 1-\left( {{A}\over {A_1}}\right) ^{{1}\over {\alpha }} \right] .$ Hence,

  $$\begin{aligned} V+{{R} \over {\gamma }}= & {} V-V_1 \left[ 1-\left( {{A}\over {A_1}}\right) ^{{1}\over {\alpha }} \right] = V-V_1+ V_1 \left( {{A}\over {A_1}}\right) ^{{1}\over {\alpha }}> V-V_1 > 0. \end{aligned}$$

  Thus, $V+{{R}\over {\gamma }} > 0$, and we find $V > -{{R}\over {\gamma }}$.

• (iii.a) Since (ii.3), we have $-{{R}\over {\gamma }} = V_1 \left[ 1-\left( {{A}\over {A_1}}\right) ^{{1}\over {\alpha }} \right] .$ Hence

  $$\begin{aligned}&V\leqslant -{{R}\over {\gamma }} \Leftrightarrow V\leqslant V_1 \left[ 1-\left( {{A}\over {A_1}}\right) ^{{1}\over {\alpha }} \right] \Leftrightarrow \left( {{A}\over {A_1}}\right) ^{{1}\over {\alpha }} \leqslant \left[ 1-{{V}\over {V_1}} \right] \\ &\quad \Leftrightarrow A \leqslant A_1 \left[ 1-{{V}\over {V_1}} \right] ^\alpha = A_0 \Leftrightarrow A \leqslant A_0. \end{aligned}$$

• (iii.b) The function $\psi$ is increasing, satisfies $\psi (0)= -\infty$ and $\psi (A_1)=0$. Therefore, $A_3 >A_1$ since $\psi (A_3)= {{V}\over {V_1}}$. Since ${{R}\over {\gamma \Phi (A)}} = {{1}\over {A_1 ^{{1}\over {\alpha }}}} \left[ {{A ^{{1}\over {\alpha }}-A_ 1 ^{{1}\over {\alpha }}}\over {\Phi (A)}} \right] V_1,$ we have

  $$\begin{aligned} {{R}\over {\gamma \Phi (A)}} \leqslant V&\Leftrightarrow \psi (A) \leqslant {{V}\over {V_1}}=\psi (A_3) \Leftrightarrow A \leqslant A_3. \end{aligned}$$

$\square$

1.5 Appendix 5: Proof of Proposition 2.3

Proof

Consider case (i). We have the following results:

• (i.a) We have $A_0 <A_1$, and hence $A\leqslant A_0 \Rightarrow A<A_1 \Leftrightarrow R<0$ (see (i) of Lemma 2.4). From (iii.a) of Lemma 2.4 we have $A\leqslant A_0 \Leftrightarrow V\leqslant -{{R}\over {\gamma }}$.

• (i.b) We have $A_1 <A_3$ (see (iii.b) of Lemma 2.4). First suppose $A_0<A<A_1$. We know that $A<A_1 \Leftrightarrow R<0$ from (i) of Lemma 2.4, and $A_0 <A \Leftrightarrow V> -{{R}\over {\gamma }}$ from (iii.a) of the same lemma. We have $-{{R}\over {\gamma }}< V \leqslant V_1 < Q(P_1, P_2, \alpha , w, \Phi , \gamma , A).$ Now suppose $A_1 \leqslant A <A_3$. If $A_1 \leqslant A \Leftrightarrow R\geqslant 0$ from (i) of Lemma 2.4 and $A<A_3 \Leftrightarrow {{R}\over {\gamma \Phi (A)}} <V$. We have $R\geqslant 0, \; {{R}\over {\gamma \Phi (A)}}<V \leqslant V_1 < Q(P_1, P_2, \alpha , w, \Phi , \gamma , A).$

• (i.c) We know that $A_3 >A_1$ (see (iii.b) of Lemma 2.4). Hence $A\geqslant A_3 \Rightarrow A>A_1 \Leftrightarrow R>0$. But $A\geqslant A_3 \Leftrightarrow V\leqslant {{R}\over {\gamma \Phi (A)}}$ (see (ii.b) of Lemma 2.4).

Apply Proposition 2.1 to get the results for each subcase.

We now consider case (ii). We distinguish two cases:

• (ii.a) $V \geqslant V_1 \left\{ 1+{{1}\over {\Phi (A_1)}}\right\} \Leftrightarrow A_2 \geqslant A_1$,

• (ii.b) $V_1<V < V_1 \left\{ 1+{{1}\over {\Phi (A_1)}}\right\} \Leftrightarrow A_1 > A_2$.

Consider case (ii.a)

(ii.a1) We have $A\leqslant A_2 \Leftrightarrow V\geqslant Q(P_1, P_2, \alpha , w, \Phi , \gamma , A)$.

(ii.a2) If $A>A_2$ then $A>A_1 \Rightarrow R>0$. Also

$$\begin{aligned} A>A_2\Leftrightarrow & {} V< Q(P_1, P_2, \alpha , w, \Phi , \gamma , A)\\ A<A_3\Leftrightarrow & {} {{R}\over {\gamma \Phi (A)}} <V. \end{aligned}$$

Apply Proposition 2.1 to get the results

• (ii.a3) If $A\geqslant A_3$ then $A>A_1 \Rightarrow R> 0$ and $V\leqslant {{R}\over {\gamma \Phi (A)}}$ (see (ii.a) of Lemma 2.4). Apply Proposition 2.1. Consider case (ii.b).

  • (ii.b1) We have $A\leqslant A_2 \Leftrightarrow V\geqslant Q(P_1, P_2, \alpha , w, \Phi , \gamma , A)$. Apply Proposition 2.1.

  • (ii.b2) We first consider the case $A_2 <A\leqslant A_1$. We have successively

    $$\begin{aligned} A\leqslant A_1\Leftrightarrow & {} R\leqslant 0 \text{(see (i) of Lemma 2.4) }\\ A>A_2\Leftrightarrow & {} V<Q(P_1, P_2, \alpha , w, \Phi , \gamma , A) \text{(see (ii.1) of Lemma 2.4) }\\ V>V_1\Rightarrow & {} V> -{{R}\over {\gamma }} \text{(see (ii.c) of Lemma 2.4) }. \end{aligned}$$

    Hence $-{{R}\over {\gamma }}<V < Q(P_1, P_2, \alpha , w, \Phi , \gamma , A).$ Apply Proposition 2.1. Now, consider the case $A_1<A<A_3$. We have successively

    $$\begin{aligned} A> A_1\Leftrightarrow & {} R>0 (\text{see}\ (\text{i})\ \text{of Lemma }2.4)\\ A>A_1\Rightarrow & {} A>A_2 \Leftrightarrow V<Q(P_1, P_2, \alpha , w, \Phi , \gamma , A) (\text{see} (\text{ii.a})\ \text{of Lemma }2.4)\\ A<A_3\Leftrightarrow & {} V> {{R}\over {\gamma \Phi (A)}}(\text{see} (\text{ii.b})\ \text{of Lemma }2.4). \end{aligned}$$

    To sum up $R>0,\; {{R}\over {\gamma \Phi (A)}}<V < Q(P_1, P_2, \alpha , w, \Phi , \gamma , A).$ Apply Proposition 2.1.

  • (ii.b3) Now assume $A\geqslant A_3$. If $A\geqslant A_3$ then $A>A_1 \Rightarrow R> 0$ and $V\leqslant {{R}\over {\gamma \Phi (A)}}$ (see (ii.b) of Lemma 2.4). Apply Proposition 2.1.

$\square$

1.6 Appendix 6: Proof of Proposition 2.6

Proof

We have $Y_2^* = {\widehat{Y}}_2 \Leftrightarrow A S_2^{* \alpha } L_2^{* 1-\alpha } = {\widehat{Y}}_2\Leftrightarrow A L_2^{* 1-\alpha } = {\widehat{Y}}_2 S_2^{* -\alpha }.$ Thus $S^* _2>0, L^* _2>0$. The problem of the producer is

$$\begin{aligned} \max \{P_1 S_1^{\alpha } L_1^{1-\alpha } - wL_1 - C(S_1) + P_2 {\widehat{Y}}_2 - w \left\{ \frac{{\widehat{Y}}_2}{A}\right\} ^{\frac{1}{(1-\alpha )}} S_2^{ -\frac{\alpha }{(1-\alpha )}} - \Phi (A) C(S_2) \end{aligned}$$

using the constraints $S_1 \geqslant 0, \; S_2 \geqslant 0, \; L_1\geqslant 0, \; S_1 +S_2 \leqslant V.$ As previously, since V is strictly larger than total land of the problem without land constraint, i.e., $S^* _1 + S^* _2 <V$, we have $S^* _1< V$ and $S^* _2<V$. We will prove that $S^* _1>0$ (and hence $L^* _1>0$). Indeed if $S^* _1=0$ then $L^* _1=0$ and $P_1{ S^* _1}^{\alpha } {L^* _1}^{1-\alpha } - wL^* _1 - C(S^* _1)=0$. But we take

$$\begin{aligned} L_1= S_1 \left[ \frac{P_1 (1-\alpha )}{w} \right] ^{\frac{1}{\alpha }},\quad S_1 = \frac{1}{\gamma } P_1 ^{\frac{1}{\alpha }}\alpha \left[ \frac{(1-\alpha )}{w} \right] ^{\frac{1-\alpha }{\alpha }} \end{aligned}$$

then $P_1 S_1^{\alpha } L_1^{1-\alpha } - wL_1 - C(S_1)= \frac{1}{2 \gamma } P^{\frac{2}{\alpha }}\alpha ^2 w^{\frac{2(\alpha -1)}{\alpha }}(1-\alpha )^{\frac{2(1-\alpha )}{\alpha }}>0$ and that is a contradiction. Hence $S^* _1>0, L^* _1>0$.

We obtain the following First-Order Conditions (FOC)

$$\begin{aligned} P_1 \alpha \left( \frac{L^{*}_1}{S^{*}_1}\right) ^{1-\alpha } - \gamma S^{*}_1 = 0 \end{aligned}$$

(15)

$$\begin{aligned} P_1 (1-\alpha ) \left( \frac{S^*_1}{L^{*}_1}\right) ^{\alpha } - w = 0 \end{aligned}$$

(16)

$$\begin{aligned} w \left\{ \frac{{\widehat{Y}}_2}{A}\right\} ^{\frac{1}{(1-\alpha )}} \frac{\alpha }{(1-\alpha )} S_2^{ -\frac{1}{(1-\alpha )}} - \Phi (A) \gamma S^{*}_2 = 0 \end{aligned}$$

(17)

From Eq. (16), we have

$$\begin{aligned} \left( \frac{S^{*}_1}{L^{*}_1}\right) ^{\alpha } = \frac{w}{P_1(1-\alpha )} \Leftrightarrow \left( \frac{L^{*}_1}{S^{*}_1}\right) ^{(1-\alpha )} = \left\{ \frac{P_1(1-\alpha )}{w}\right\} ^{\frac{(1-\alpha )}{\alpha }}. \end{aligned}$$

(18)

From Eqs. (15) and (18), we get

$$\begin{aligned} P_1^{\frac{1}{\alpha }} \alpha \left\{ \frac{1-\alpha }{w}\right\} ^{\frac{(1-\alpha )}{\alpha }} = \gamma S^{*}_1. \end{aligned}$$

(19)

From Eq. (17), we obtain

$$\begin{aligned} \frac{\alpha }{(1-\alpha )} w \left\{ \frac{{\widehat{Y}}_2}{A}\right\} ^{\frac{1}{(1-\alpha )}} S_2^{ -\frac{1}{(1-\alpha )}} = \gamma \Phi (A) S^{*}_2. \end{aligned}$$

(20)

From Eqs. (19) and (20), we obtain

$$\begin{aligned} P_1^{\frac{1}{\alpha }} \alpha \left\{ \frac{1-\alpha }{w} \right\} ^{\frac{(1-\alpha )}{\alpha }}= {} \gamma S^{*}_1 \end{aligned}$$

(21)

$$\begin{aligned} \frac{\alpha }{(1-\alpha )} w \left\{ \frac{{\widehat{Y}}_2}{A} \right\} ^{\frac{1}{(1-\alpha )}} S_2^{ -\frac{1}{(1-\alpha )}}= {} \gamma \Phi (A) S^{*}_2. \end{aligned}$$

(22)

From Eqs. (21) and (14), we get

$$\begin{aligned} \frac{\alpha }{\gamma } P_1^{\frac{1}{\alpha }} \left\{ \frac{1-\alpha }{w}\right\} ^{\frac{(1-\alpha )}{\alpha }}= {} S^{*}_1\end{aligned}$$

(23)

$$\begin{aligned} \frac{\alpha }{(1-\alpha )} \frac{ w }{\gamma \Phi (A)} \left\{ \frac{{\widehat{Y}}_2}{A}\right\} ^{\frac{1}{(1-\alpha )}}= {} S_2^{* \frac{(2-\alpha )}{(1-\alpha )}}. \end{aligned}$$

(24)

Summing up:

$$\begin{aligned}&{\textit{land for conventional production}}, S^{*}_{1} = \frac{\alpha }{\gamma } P_1^{\frac{1}{\alpha }} \left\{ \frac{1-\alpha }{w}\right\} ^{\frac{(1-\alpha )}{\alpha }},\\ &{\textit{labor for conventional production}}, L^{*}_{1}= \frac{\alpha }{\gamma } P_1^{\frac{2}{\alpha }} \left\{ \frac{1-\alpha }{w}\right\} ^{\frac{(2-\alpha )}{\alpha }},\\ &{\textit{land for organic production}}, S^{*}_{2}= \left\{ \frac{\alpha }{(1-\alpha )} \frac{w}{\gamma \Phi (A)}\right\} ^{\frac{1-\alpha }{(2-\alpha )}} \left\{ \frac{{\widehat{Y}}_2}{A}\right\} ^{\frac{1}{(2-\alpha )}},\\ &{\textit{labor for organic production}}, L^{*}_{2}= \left\{ \frac{\alpha }{(1-\alpha )} \frac{w}{\gamma \Phi (A)}\right\} ^{-\frac{\alpha }{(2-\alpha )}} \left\{ \frac{{\widehat{Y}}_2}{A}\right\} ^{\frac{2}{(2-\alpha )}}. \end{aligned}$$

Observe that actually $S^{*}_{1} + S^{*}_{2} < V$. Hence, the values $S^{*}_{1}, \; L^{*}_{1}, \; S^{*}_{2}, \; L^{*}_{2}$ are optimal. $\square$