<p>I found BakerStreet's answer useful to intuitively make conclusions about the model, however I was then able to explicitly solve for the household's agents' optimal time allocations myself with a trick:</p>
<p>Total consumption is given by:</p>
<p><span class="math-container">$c = \frac{1}{3} [2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]$</span></p>
<p><span class="math-container">$\ln{c} = - \ln{3} + \ln{[2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]}$</span></p>
<p>Substituting this expression for <span class="math-container">$\ln c$</span> into <span class="math-container">$U$</span>, we get the following program:</p>
<p><span class="math-container">$\max U(N^F,N^M,V^F,V^M,l^F,l^M) = - \ln{3} + \ln{[2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]} + \ln l^F + \ln l^M$</span></p>
<p>subject to both agents' time constraints</p>
<p><span class="math-container">$l^F + V^F + N^F = 1$</span></p>
<p><span class="math-container">$l^M + V^M + N^M = 1$</span></p>
<p>We form the corresponding Lagrangian with two Lagrange multipliers: a female lambda <span class="math-container">$\lambda^F$</span> and a male lambda <span class="math-container">$\lambda^M$</span>:</p>
<p><span class="math-container">$\mathcal{L} = - \ln{3} + \ln{[2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]} + \ln l^F + \ln l^M + \lambda^F (1-l^F-V^F-N^F) + \lambda^M (1-l^M-V^M-N^M)$</span></p>
<p>We get the following first order conditions:</p>
<p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial N^F} = \frac{2w^F}{A} - \lambda^F = 0 \implies \lambda^F = \frac{2w^F}{A}$</span></p>
<p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial l^F} = \frac{1}{l^F} - \lambda^F = 0 \implies \lambda^F = \frac{1}{l^F}$</span></p>
<p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial V^F} = \frac{\frac{1}{2} (V^F)^{-\frac{1}{2}} (V^M)^{\frac{1}{2}}}{A} - \lambda^F = 0 \implies \lambda^F = \frac{\frac{1}{2} (V^F)^{-\frac{1}{2}} (V^M)^{\frac{1}{2}}}{A}$</span></p>
<p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial N^M} = \frac{2w^M}{A} - \lambda^M = 0 \implies \lambda^M = \frac{2w^M}{A}$</span></p>
<p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial l^M} = \frac{1}{l^M} - \lambda^M = 0 \implies \lambda^M = \frac{1}{l^M}$</span></p>
<p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial V^M} = \frac{\frac{1}{2} (V^F)^{\frac{1}{2}} (V^M)^{-\frac{1}{2}}}{A} - \lambda^M = 0 \implies \lambda^M = \frac{\frac{1}{2} (V^F)^{\frac{1}{2}} (V^M)^{-\frac{1}{2}}}{A}$</span></p>
<p>where <span class="math-container">$A:= 2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}$</span> as BakerStreet defined it.</p>
<p>Now we calculate the ratios <span class="math-container">$\frac{\lambda^F}{\lambda^M}$</span> by pairing each female first order condition with its male dual:</p>
<p><span class="math-container">$\frac{\lambda^F}{\lambda^M} = \frac{w^F}{w^M}$</span></p>
<p><span class="math-container">$\frac{\lambda^F}{\lambda^M} = \frac{l^M}{l^F} \implies l^M = \frac{w^F}{w^M} l^F$</span></p>
<p><span class="math-container">$\frac{\lambda^F}{\lambda^M} = \frac{V^M}{V^F} \implies V^M = \frac{w^F}{w^M} V^F$</span></p>
<p>These intergender conditions also imply that</p>
<p><span class="math-container">$N^M = 1 - \frac{w^F}{w^M} l^F - \frac{w^F}{w^M} V^F$</span></p>
<p>Plugging these equations and the female time constraint into our objective function, we reduce it to <span class="math-container">$2$</span> variables:</p>
<p><span class="math-container">$U(N^F,V^F) = \ln[2w^FN^F+2w^M-2w^F(1-N^F-V^F)-2w^FV^F+2\Pi+(\frac{w^F}{w^M})^{\frac{1}{2}}V^F] + 2 \ln(1-N^F-V^F) + [\ln(\frac{w^F}{w^M}) - \ln 3]$</span></p>
<p>Notice <span class="math-container">$A = 2w^FN^F+2w^M-2w^F(1-N^F-V^F)-2w^FV^F+2\Pi+(\frac{w^F}{w^M})^{\frac{1}{2}}V^F$</span>.</p>
<p>Our new first order conditions are:</p>
<p><span class="math-container">$\frac{\partial U}{\partial N^F} = \frac{4w^F}{A} - \frac{2}{1-N^F-V^F}$</span></p>
<p><span class="math-container">$\frac{\partial U}{\partial V^F} = \frac{(\frac{w^F}{w^M})^{\frac{1}{2}}}{A} - \frac{2}{1-N^F-V^F}$</span></p>
<p>Doing some algebra, we transform our first order conditions into the following system of <span class="math-container">$2$</span> linear equations in <span class="math-container">$2$</span> variables <span class="math-container">$(N^F,V^F)$</span></p>
<p><span class="math-container">$(\frac{w^F}{w^M})^\frac{1}{2} V^F = 4 w^F - 2 w^M - 2 \Pi - 6 w^F N^F$</span></p>
<p><span class="math-container">$(\frac{w^F}{w^M})^\frac{1}{2} V^F = (\frac{w^F}{w^M})^\frac{1}{2} + 4 w^F - 4 w^M - 4 \Pi + [(\frac{w^F}{w^M})^\frac{1}{2} - 8 w^F] N^F$</span></p>
<p>Since the left hand sides are equal, equating both right hand sides we solve for <span class="math-container">$N^F$</span> and get:</p>
<p><span class="math-container">$N^F = \frac{2w^M+2\Pi-(\frac{w^F}{w^M})^\frac{1}{2}}{(\frac{w^F}{w^M})^\frac{1}{2} - 2 w^F}$</span></p>
<p>Plugging into either of both equations, we explicitly get <span class="math-container">$V^F$</span>.</p>
<p>Plugging <span class="math-container">$N^F,V^F$</span> into the female time constraint, we get <span class="math-container">$l^F$</span>.</p>
<p>Plugging <span class="math-container">$N^F,V^F$</span> into the intergender conditions, we get <span class="math-container">$N^M,V^M$</span>.</p>
<p>Plugging <span class="math-container">$N^M,V^M$</span> into the male time constraint, we get <span class="math-container">$l^M$</span>.</p>
<p>The only difference in optimal domestic labor time assignations depends on the ratio of the agents' wages in the paid labor market, as per BakerStreet's intuition.</p>

CC BY-SA 4.0
2023-05-05T13:59:32.397Z
Thank youy very much. Your further elaboration derivs explicitely the labor supplies.
BakerStreet
CC BY-SA 4.0
2023-05-05T14:13:54.077Z
Actually, the model is a little complicated from a formal point of view, but it gives mathematical form to what any sensible person can think: if your salary is greater than mine, it is better, *ceteris paribus* , that you work more than me in non-domestic job. +1 for your solution.
BakerStreet
CC BY-SA 4.0
2023-05-05T13:56:43.600Z
2023-05-05T17:41:13.493Z
2023-05-05T17:41:13.493Z
Nicolas Torres
<p>In principle, using Lagrange multipliers instead of substitution should be the same.  <strong>Lagrange multipliers</strong> are important when the <strong>constraint functions can't be explicitated</strong>, otherwise they are not necessary. So, I think that this isn't the problem.</p>
<p>The problem is how to obtain something useful and meaningful from the first order conditions.</p>
<p>First, I observe that there isn't any difference between <strong>female and male work</strong> in the problem, the two <strong>enter the problem in a symmetric way</strong>. The only differences in the result of maximization can come from different wages, <span class="math-container">$w^F$</span> and <span class="math-container">$w^M$</span>.</p>
<p>This suggests that could be meaningful to <strong>find the labor supplies in dependance on the wages</strong>, actually nothing unusual (I'm re-discovering the labor supply function...).</p>
<p><span class="math-container">$\;$</span></p>
<p>I propose the following way.</p>
<p>Observe that the first term of the equations for the derivatives has  the same denominator, that I call <span class="math-container">$A$</span> to simplify the calculations. That is, I set:</p>
<p><span class="math-container">${2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}}=A$</span>.</p>
<p>Therefore, the first order conditions can be re-written as</p>
<p><span class="math-container">$\frac{\partial U}{\partial N^F} = \frac{2 w^F}{A} - \frac{1}{1-N^F-V^F} = 0\qquad \;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$</span></p>
<p><span class="math-container">$\frac{\partial U}{\partial N^M} = \frac{2 w^M}{A} - \frac{1}{1-N^M-V^M} = 0\;\;\;\;\;\;\;\;\;\;\;\;\qquad (2)$</span></p>
<p><span class="math-container">$\frac{\partial U}{\partial V^F} = \frac{\frac{1}{2} (V^F)^{-\frac{1}{2}} (V^M)^{\frac{1}{2}}}{A} - \frac{1}{1-N^F-V^F} = 0\;\qquad (3)$</span></p>
<p><span class="math-container">$\frac{\partial U}{\partial V^M} = \frac{\frac{1}{2} (V^F)^{\frac{1}{2}} (V^M)^{-\frac{1}{2}}}{A} - \frac{1}{1-N^M-V^M} = 0\qquad (4)$</span></p>
<p>From <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> we have:</p>
<p><span class="math-container">$2 w^F = \frac{A}{1-N^F-V^F}  $</span></p>
<p><span class="math-container">$2 w^M = \frac{A}{1-N^M-V^M}  $</span></p>
<p>and taking the ratio</p>
<p><span class="math-container">$\frac{w^F}{w^M}=\frac{1-N^M-V^M}{1-N^F-V^F}.  $</span></p>
<p>Remembering that</p>
<p><span class="math-container">$l^i + V^i + N^i = 1$</span></p>
<p>we have</p>
<p><span class="math-container">$\frac{w^F}{w^M}=\frac{l^M}{l^F} \;\;\;\;\;\;\;\;\;\;\;\;\qquad (5) $</span></p>
<p>Expression <span class="math-container">$(5)$</span> says us that the <strong>amounts of leisure of the female and the male worker depends (inversely) on the wages</strong> (their ratios must be equal), in particular if the wages <span class="math-container">$w^F$</span> and <span class="math-container">$w^M$</span> are equal, the amount of leisure of the two workers will be equal.</p>
<p><span class="math-container">$\;$</span></p>
<p>How the total amount of work is <strong>distributed among paid and domestic work?</strong></p>
<p>Consider the first order conditions <span class="math-container">$(3)$</span> and <span class="math-container">$(4)$</span>. Taking the ratio we can write</p>
<p><span class="math-container">$ \frac{\frac{1}{2} (V^F)^{-\frac{1}{2}}(V^M)^{\frac{1}{2}}}{\frac{1}{2} (V^F)^{\frac{1}{2}}(V^M)^{-\frac{1}{2}}}=\frac{1-N^M-V^M}{1-N^F-V^F}$</span></p>
<p>that is</p>
<p><span class="math-container">$\frac{(V^M)^{1/4}}{(V^F)^{1/4}}=\frac{1-N^M-V^M}{1-N^F-V^F}$</span></p>
<p>which can be written as</p>
<p><span class="math-container">$\frac{V^M}{V^F}=\left(\frac{l^M}{l^F}\right)^4. \;\;\;\;\;\;\;\;\;\;\;\;\qquad (6)$</span></p>
<p>Equation <span class="math-container">$(6)$</span> says us that the ratio of domestic work of male and female workers depends (positively) on the ratio of leisure, which in turn depends (negatively) on the ratio of wages.</p>
<p>That is, a member of the household <strong>will have a greater amount of non-paid, domestic, labor as lower is their wage.</strong></p>
<p><span class="math-container">$$***$$</span></p>
<p>In conclusion, we can say the supplies, domestic and paid, of labor of both members of the household, and leisure, are linked to the relevant wages.</p>
<p>If the female and male wages are equal, all labor supplies and leisure will be equal.</p>
<p>Nothing surprising, as it confirms our first intuitive impression, as the model is symmetric with  respect to the work of both members of the household, and the only possible differences lie in the wages.</p>
<p>Of course, further or different elaborations could be possible, but this way I think we have already meaningful results.</p>

CC BY-SA 4.0
2023-05-05T13:59:21.273Z
Thank you for the answer! I was able to explicitly solve for the time assignations myself, however I found your answer useful for making conclusions and saving some time in the calculations.
Nicolas Torres
CC BY-SA 4.0
2023-05-03T13:00:09.327Z
2023-05-03T15:53:11.353Z
2023-05-03T15:53:11.353Z
BakerStreet
<p>Consider a household consisting of a woman and a man, with preferences over leisure and consumption given by:</p>
<p><span class="math-container">$U(\overrightarrow{c},\overrightarrow{l}) = \ln{c} + \ln{l^F} + \ln{l^M}$</span></p>
<p>where <span class="math-container">$\overrightarrow{l} = (l^F,l^M)$</span> is the vector of leisure times for the female and male, respectively.</p>
<p>On the other hand, consumption <span class="math-container">$c$</span> is defined by the household's preferences over consumption vectors <span class="math-container">$\overrightarrow{c} = (c^N,c^V)$</span> given as</p>
<p><span class="math-container">$c = \frac{2}{3} c^N + \frac{1}{3} c^V$</span></p>
<p>where <span class="math-container">$c^N$</span> is market goods (which I'd think of as groceries) and <span class="math-container">$c^V$</span> is consumption generated by domestic labor (which I'd think of as pure satisfaction of having a clean home)</p>
<p>Consumption of domestic labor is given by (a kind of production function)</p>
<p><span class="math-container">$c^V = (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}$</span></p>
<p>where <span class="math-container">$V = (V^F,V^M)$</span> are the domestic labor times for the female and male, respectively.</p>
<p>Consumption of market goods is given by the budget constraint</p>
<p><span class="math-container">$c^N = w^F N^F + w^M N^M + \Pi$</span></p>
<p>where <span class="math-container">$N = (N^F,N^M)$</span> are the paid labor times for the female and male, respectively; <span class="math-container">$\Pi$</span> is the firm's profits, and <span class="math-container">$(w^F,w^M)$</span> are the wages for the female and male, respectively.</p>
<p>Here both individuals have a time endowment of <span class="math-container">$1$</span> unit:</p>
<p><span class="math-container">$l^i + V^i + N^i = 1$</span></p>
<p>Find the female's and male's paid labor supply and domestic labor times and compare.</p>
<hr />
<p>Here is what I have done:</p>
<p>Plugging <span class="math-container">$c^N, c^V$</span> into the preferences for <span class="math-container">$c$</span>,</p>
<p><span class="math-container">$c = \frac{1}{3} [2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]$</span></p>
<p><span class="math-container">$\ln{c} = - \ln{3} + \ln{[2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]}$</span></p>
<p>From the time constraint I can eliminate the leisure variables by rewriting <span class="math-container">$l^i = 1 - N^i - V^i$</span></p>
<p>Plugging my new expressions for <span class="math-container">$c$</span> and the <span class="math-container">$l^i$</span> into the utility function:</p>
<p><span class="math-container">$U(N^F,N^M,V^F,V^M) = - \ln{3} + \ln{[2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]} + \ln{(1-N^F-V^F)} + \ln{(1-N^M-V^M)}$</span></p>
<p>From here I'd get the first order conditions by differentiating with respect to each of the four variables and setting <span class="math-container">$= 0$</span>.</p>
<p><span class="math-container">$\frac{\partial U}{\partial N^F} = \frac{2 w^F}{2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}} - \frac{1}{1-N^F-V^F} = 0$</span></p>
<p><span class="math-container">$\frac{\partial U}{\partial N^M} = \frac{2 w^M}{2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}} - \frac{1}{1-N^M-V^M} = 0$</span></p>
<p><span class="math-container">$\frac{\partial U}{\partial V^F} = \frac{\frac{1}{2} (V^F)^{-\frac{1}{2}} (V^M)^{\frac{1}{2}}}{2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}} - \frac{1}{1-N^F-V^F} = 0$</span></p>
<p><span class="math-container">$\frac{\partial U}{\partial V^M} = \frac{\frac{1}{2} (V^F)^{\frac{1}{2}} (V^M)^{-\frac{1}{2}}}{2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}} - \frac{1}{1-N^M-V^M} = 0$</span></p>
<p>However, solving this system of <span class="math-container">$4$</span> equations seems rather complicated.</p>
<p>I tried pairing each two equations with the same negative term and get an equation from each pair, to reduce it to a system of two equations.</p>
<p>However, when I try to solve the new system, I get <span class="math-container">$w^F w^M = \frac{1}{16}$</span>, which doesn't seem very helpful or even make sense.</p>
<p>Is this method of substituting the consumption equations at the start wrong here, and I should rather try to form a (very long) Lagrangian? Is there a more clever way to solve this problem?</p>
<p>I think I may be able to use marginal rates of substitution but this seems very confusing to me as in this situation, there are now <span class="math-container">$3$</span> possible ways the household agents could spend their time, rather than <span class="math-container">$2$</span>; aside from the fact that the household consists of <span class="math-container">$2$</span> agents rather than one.</p>

CC BY-SA 4.0
2023-05-02T12:30:39.140Z
I do not have time to write whole answer but intuitively I think if you use lagrangian It will be easier, also if you have system of equations like this you can use cramer's rule, you have lot of variables there which makes it pain either way
1muflon1
CC BY-SA 4.0
2023-05-02T12:36:13.407Z
@1muflon1 Thank you for your feedback! How can I apply Cramer’s rule when the equations are nonlinear? I’ll try to solve the Lagrangian with the $5$ lambdas I think (one for each of $c,c^N,c^V,l^F,l^M$) and see if it’s clearer.
Nicolas Torres
CC BY-SA 4.0
2023-05-02T14:47:09.490Z
couldn't you just linearize the system and apply it then? I am not sure now to be honest
1muflon1
<macroeconomics><microeconomics><utility><labor-economics><optimization>
Utility maximization for a household consisting of a woman and a man, with gender discrimination
Nicolas Torres
# Utility maximization for a household consisting of a woman and a man, with gender discrimination

(Asked by: username_0 on 2023-05-02)

Consider a household consisting of a woman and a man, with preferences over leisure and consumption given by:

$U(\overrightarrow{c},\overrightarrow{l}) = \ln{c} + \ln{l^F} + \ln{l^M}$

where $\overrightarrow{l} = (l^F,l^M)$ is the vector of leisure times for the female and male, respectively.

On the other hand, consumption $c$ is defined by the household's preferences over consumption vectors $\overrightarrow{c} = (c^N,c^V)$ given as

$c = \frac{2}{3} c^N + \frac{1}{3} c^V$

where $c^N$ is market goods (which I'd think of as groceries) and $c^V$ is consumption generated by domestic labor (which I'd think of as pure satisfaction of having a clean home)

Consumption of domestic labor is given by (a kind of production function)

$c^V = (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}$

where $V = (V^F,V^M)$ are the domestic labor times for the female and male, respectively.

Consumption of market goods is given by the budget constraint

$c^N = w^F N^F + w^M N^M + \Pi$

where $N = (N^F,N^M)$ are the paid labor times for the female and male, respectively; $\Pi$ is the firm's profits, and $(w^F,w^M)$ are the wages for the female and male, respectively.

Here both individuals have a time endowment of $1$ unit:

$l^i + V^i + N^i = 1$

Find the female's and male's paid labor supply and domestic labor times and compare.

---

Here is what I have done:

Plugging $c^N, c^V$ into the preferences for $c$,

$c = \frac{1}{3} [2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]$

$\ln{c} = - \ln{3} + \ln{[2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]}$

From the time constraint I can eliminate the leisure variables by rewriting $l^i = 1 - N^i - V^i$

Plugging my new expressions for $c$ and the $l^i$ into the utility function:

$U(N^F,N^M,V^F,V^M) = - \ln{3} + \ln{[2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]} + \ln{(1-N^F-V^F)} + \ln{(1-N^M-V^M)}$

From here I'd get the first order conditions by differentiating with respect to each of the four variables and setting $= 0$.

$\frac{\partial U}{\partial N^F} = \frac{2 w^F}{2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}} - \frac{1}{1-N^F-V^F} = 0$

$\frac{\partial U}{\partial N^M} = \frac{2 w^M}{2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}} - \frac{1}{1-N^M-V^M} = 0$

$\frac{\partial U}{\partial V^F} = \frac{\frac{1}{2} (V^F)^{-\frac{1}{2}} (V^M)^{\frac{1}{2}}}{2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}} - \frac{1}{1-N^F-V^F} = 0$

$\frac{\partial U}{\partial V^M} = \frac{\frac{1}{2} (V^F)^{\frac{1}{2}} (V^M)^{-\frac{1}{2}}}{2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}} - \frac{1}{1-N^M-V^M} = 0$

However, solving this system of $4$ equations seems rather complicated.

I tried pairing each two equations with the same negative term and get an equation from each pair, to reduce it to a system of two equations.

However, when I try to solve the new system, I get $w^F w^M = \frac{1}{16}$, which doesn't seem very helpful or even make sense.

Is this method of substituting the consumption equations at the start wrong here, and I should rather try to form a (very long) Lagrangian? Is there a more clever way to solve this problem?

I think I may be able to use marginal rates of substitution but this seems very confusing to me as in this situation, there are now $3$ possible ways the household agents could spend their time, rather than $2$; aside from the fact that the household consists of $2$ agents rather than one.

**Tags:** macroeconomics, microeconomics, utility, labor-economics, optimization
> **Comments:**
> **username_1**: I do not have time to write whole answer but intuitively I think if you use lagrangian It will be easier, also if you have system of equations like this you can use cramer's rule, you have lot of variables there which makes it pain either way  
> **username_0**: @1muflon1 Thank you for your feedback! How can I apply Cramer’s rule when the equations are nonlinear? I’ll try to solve the Lagrangian with the $5$ lambdas I think (one for each of $c,c^N,c^V,l^F,l^M$) and see if it’s clearer.  
> **username_1**: couldn't you just linearize the system and apply it then? I am not sure now to be honest  

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## 2 answers

### Answer by username_2 on 2023-05-03. Score: 3 | Accepted answer

In principle, using Lagrange multipliers instead of substitution should be the same. **Lagrange multipliers** are important when the **constraint functions can't be explicitated**, otherwise they are not necessary. So, I think that this isn't the problem.

The problem is how to obtain something useful and meaningful from the first order conditions.

First, I observe that there isn't any difference between **female and male work** in the problem, the two **enter the problem in a symmetric way**. The only differences in the result of maximization can come from different wages, $w^F$ and $w^M$.

This suggests that could be meaningful to **find the labor supplies in dependance on the wages**, actually nothing unusual (I'm re-discovering the labor supply function...).

$\;$

I propose the following way.

Observe that the first term of the equations for the derivatives has the same denominator, that I call $A$ to simplify the calculations. That is, I set:

${2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}}=A$.

Therefore, the first order conditions can be re-written as

$\frac{\partial U}{\partial N^F} = \frac{2 w^F}{A} - \frac{1}{1-N^F-V^F} = 0\qquad \;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$

$\frac{\partial U}{\partial N^M} = \frac{2 w^M}{A} - \frac{1}{1-N^M-V^M} = 0\;\;\;\;\;\;\;\;\;\;\;\;\qquad (2)$

$\frac{\partial U}{\partial V^F} = \frac{\frac{1}{2} (V^F)^{-\frac{1}{2}} (V^M)^{\frac{1}{2}}}{A} - \frac{1}{1-N^F-V^F} = 0\;\qquad (3)$

$\frac{\partial U}{\partial V^M} = \frac{\frac{1}{2} (V^F)^{\frac{1}{2}} (V^M)^{-\frac{1}{2}}}{A} - \frac{1}{1-N^M-V^M} = 0\qquad (4)$

From $(1)$ and $(2)$ we have:

$2 w^F = \frac{A}{1-N^F-V^F} $

$2 w^M = \frac{A}{1-N^M-V^M} $

and taking the ratio

$\frac{w^F}{w^M}=\frac{1-N^M-V^M}{1-N^F-V^F}. $

Remembering that

$l^i + V^i + N^i = 1$

we have

$\frac{w^F}{w^M}=\frac{l^M}{l^F} \;\;\;\;\;\;\;\;\;\;\;\;\qquad (5) $

Expression $(5)$ says us that the **amounts of leisure of the female and the male worker depends (inversely) on the wages** (their ratios must be equal), in particular if the wages $w^F$ and $w^M$ are equal, the amount of leisure of the two workers will be equal.

$\;$

How the total amount of work is **distributed among paid and domestic work?**

Consider the first order conditions $(3)$ and $(4)$. Taking the ratio we can write

$ \frac{\frac{1}{2} (V^F)^{-\frac{1}{2}}(V^M)^{\frac{1}{2}}}{\frac{1}{2} (V^F)^{\frac{1}{2}}(V^M)^{-\frac{1}{2}}}=\frac{1-N^M-V^M}{1-N^F-V^F}$

that is

$\frac{(V^M)^{1/4}}{(V^F)^{1/4}}=\frac{1-N^M-V^M}{1-N^F-V^F}$

which can be written as

$\frac{V^M}{V^F}=\left(\frac{l^M}{l^F}\right)^4. \;\;\;\;\;\;\;\;\;\;\;\;\qquad (6)$

Equation $(6)$ says us that the ratio of domestic work of male and female workers depends (positively) on the ratio of leisure, which in turn depends (negatively) on the ratio of wages.

That is, a member of the household **will have a greater amount of non-paid, domestic, labor as lower is their wage.**

$$\*\*\*$$

In conclusion, we can say the supplies, domestic and paid, of labor of both members of the household, and leisure, are linked to the relevant wages.

If the female and male wages are equal, all labor supplies and leisure will be equal.

Nothing surprising, as it confirms our first intuitive impression, as the model is symmetric with respect to the work of both members of the household, and the only possible differences lie in the wages.

Of course, further or different elaborations could be possible, but this way I think we have already meaningful results.

> **Comments:**
> **username_0**: Thank you for the answer! I was able to explicitly solve for the time assignations myself, however I found your answer useful for making conclusions and saving some time in the calculations.  

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### Answer by username_0 on 2023-05-05. Score: 1

I found BakerStreet's answer useful to intuitively make conclusions about the model, however I was then able to explicitly solve for the household's agents' optimal time allocations myself with a trick:

Total consumption is given by:

$c = \frac{1}{3} [2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]$

$\ln{c} = - \ln{3} + \ln{[2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]}$

Substituting this expression for $\ln c$ into $U$, we get the following program:

$\max U(N^F,N^M,V^F,V^M,l^F,l^M) = - \ln{3} + \ln{[2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]} + \ln l^F + \ln l^M$

subject to both agents' time constraints

$l^F + V^F + N^F = 1$

$l^M + V^M + N^M = 1$

We form the corresponding Lagrangian with two Lagrange multipliers: a female lambda $\lambda^F$ and a male lambda $\lambda^M$:

$\mathcal{L} = - \ln{3} + \ln{[2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}]} + \ln l^F + \ln l^M + \lambda^F (1-l^F-V^F-N^F) + \lambda^M (1-l^M-V^M-N^M)$

We get the following first order conditions:

$\frac{\partial \mathcal{L}}{\partial N^F} = \frac{2w^F}{A} - \lambda^F = 0 \implies \lambda^F = \frac{2w^F}{A}$

$\frac{\partial \mathcal{L}}{\partial l^F} = \frac{1}{l^F} - \lambda^F = 0 \implies \lambda^F = \frac{1}{l^F}$

$\frac{\partial \mathcal{L}}{\partial V^F} = \frac{\frac{1}{2} (V^F)^{-\frac{1}{2}} (V^M)^{\frac{1}{2}}}{A} - \lambda^F = 0 \implies \lambda^F = \frac{\frac{1}{2} (V^F)^{-\frac{1}{2}} (V^M)^{\frac{1}{2}}}{A}$

$\frac{\partial \mathcal{L}}{\partial N^M} = \frac{2w^M}{A} - \lambda^M = 0 \implies \lambda^M = \frac{2w^M}{A}$

$\frac{\partial \mathcal{L}}{\partial l^M} = \frac{1}{l^M} - \lambda^M = 0 \implies \lambda^M = \frac{1}{l^M}$

$\frac{\partial \mathcal{L}}{\partial V^M} = \frac{\frac{1}{2} (V^F)^{\frac{1}{2}} (V^M)^{-\frac{1}{2}}}{A} - \lambda^M = 0 \implies \lambda^M = \frac{\frac{1}{2} (V^F)^{\frac{1}{2}} (V^M)^{-\frac{1}{2}}}{A}$

where $A:= 2w^F N^F + 2w^M N^M + 2 \Pi + (V^F)^{\frac{1}{2}} (V^M)^{\frac{1}{2}}$ as BakerStreet defined it.

Now we calculate the ratios $\frac{\lambda^F}{\lambda^M}$ by pairing each female first order condition with its male dual:

$\frac{\lambda^F}{\lambda^M} = \frac{w^F}{w^M}$

$\frac{\lambda^F}{\lambda^M} = \frac{l^M}{l^F} \implies l^M = \frac{w^F}{w^M} l^F$

$\frac{\lambda^F}{\lambda^M} = \frac{V^M}{V^F} \implies V^M = \frac{w^F}{w^M} V^F$

These intergender conditions also imply that

$N^M = 1 - \frac{w^F}{w^M} l^F - \frac{w^F}{w^M} V^F$

Plugging these equations and the female time constraint into our objective function, we reduce it to $2$ variables:

$U(N^F,V^F) = \ln[2w^FN^F+2w^M-2w^F(1-N^F-V^F)-2w^FV^F+2\Pi+(\frac{w^F}{w^M})^{\frac{1}{2}}V^F] + 2 \ln(1-N^F-V^F) + [\ln(\frac{w^F}{w^M}) - \ln 3]$

Notice $A = 2w^FN^F+2w^M-2w^F(1-N^F-V^F)-2w^FV^F+2\Pi+(\frac{w^F}{w^M})^{\frac{1}{2}}V^F$.

Our new first order conditions are:

$\frac{\partial U}{\partial N^F} = \frac{4w^F}{A} - \frac{2}{1-N^F-V^F}$

$\frac{\partial U}{\partial V^F} = \frac{(\frac{w^F}{w^M})^{\frac{1}{2}}}{A} - \frac{2}{1-N^F-V^F}$

Doing some algebra, we transform our first order conditions into the following system of $2$ linear equations in $2$ variables $(N^F,V^F)$

$(\frac{w^F}{w^M})^\frac{1}{2} V^F = 4 w^F - 2 w^M - 2 \Pi - 6 w^F N^F$

$(\frac{w^F}{w^M})^\frac{1}{2} V^F = (\frac{w^F}{w^M})^\frac{1}{2} + 4 w^F - 4 w^M - 4 \Pi + [(\frac{w^F}{w^M})^\frac{1}{2} - 8 w^F] N^F$

Since the left hand sides are equal, equating both right hand sides we solve for $N^F$ and get:

$N^F = \frac{2w^M+2\Pi-(\frac{w^F}{w^M})^\frac{1}{2}}{(\frac{w^F}{w^M})^\frac{1}{2} - 2 w^F}$

Plugging into either of both equations, we explicitly get $V^F$.

Plugging $N^F,V^F$ into the female time constraint, we get $l^F$.

Plugging $N^F,V^F$ into the intergender conditions, we get $N^M,V^M$.

Plugging $N^M,V^M$ into the male time constraint, we get $l^M$.

The only difference in optimal domestic labor time assignations depends on the ratio of the agents' wages in the paid labor market, as per BakerStreet's intuition.

> **Comments:**
> **username_2**: Thank youy very much. Your further elaboration derivs explicitely the labor supplies.  
> **username_2**: Actually, the model is a little complicated from a formal point of view, but it gives mathematical form to what any sensible person can think: if your salary is greater than mine, it is better, *ceteris paribus* , that you work more than me in non-domestic job. +1 for your solution.  

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