<p>Seems like you are assuming there will be an interior solution.
$$
(a-2b x-c^X)w-by = 0
$$
only needs to hold if the value of $x$ is positive. If it is not, but $x = 0$, the lowest possible amount, then 
$$
(a-2b x-c^X)w-by &lt; 0
$$
is not a contradiction. Further decreasing $x$ would increase utility, but it is not possible. Therefore in the case when $w = 0$ and
$$
(a-2b x-c^X)w-by = -by
$$
you do not necessarily have $y = 0$.</p>

CC BY-SA 4.0
2018-08-09T23:54:11.743Z
2018-08-09T23:54:11.743Z
Giskard
<p>Similar to the corner solution argument by denesp:  </p>

<p>When we solve Max U(x,y) when w = 0, we are selecting x and y such that U is maximized. But, w = 0, implies that U is decreasing in x. Thus, we will choose the smallest possible value of x. For any value of y we can select, a non-zero value of x will lower utility.  </p>

CC BY-SA 4.0
2018-08-10T17:50:27.783Z
2018-08-10T17:50:27.783Z
erik
<p>I'm trying to solve the optimal production $\{x,y\}$ for a risk neutral agent with weight $w$ in firm $X$ and weight $1-w$ in firm $Y$. Each firm has marginal cost $c^X$ and $c^Y$ respectively. The firms face a linear demand where $P(Q)=a-b Q$ and the total production of the economy $Q=x+y$. This risk neutral agent maximizes profits so his/her utility will be:</p>

<p>$U(x,y)=w(x(a-b(x+y)-c^X))+(1-w)(y(a-b(x+y)-c^Y))$</p>

<p>If I take first order conditions to maximize this utility I get:</p>

<p>$(a-2b x-c^X)w-by=0$</p>

<p>$(a-2b y-c^Y)(1-w)-bx=0$</p>

<p>Which solves to:</p>

<p>$x=\frac{(1-w)(2 c^X w-c^Y+a(1-2 w))}{b(1-2w)^2}$</p>

<p>$y=\frac{w(2 c^Y(1- w)-c^X-a(1-2 w))}{b(1-2w)^2}$</p>

<p>Assuming this is all correct, I don't understand why when $w=0$, then $y=0$!!! and $x=\frac{a-c^Y}{b}$ maxing the utility $U(x,y)=0$.</p>

<p>I think this doesn't make sense and can't be the optimum because having $x=0$ and $y=\frac{a-c^Y}{2b}$ (monopoly production) would give $U(x,y)=\frac{(a-c)^2
}{2b}&gt;0$</p>

<p>I must have something wrong, the derivatives and solutions are correct, does anyone see what I'm missing here?</p>

<production-function><productivity><profit-maximization>
2 firms production decision for one agent
NinjaCowAndForks
# 2 firms production decision for one agent

(Asked by: username_0 on 2018-08-09)

I'm trying to solve the optimal production $\{x,y\}$ for a risk neutral agent with weight $w$ in firm $X$ and weight $1-w$ in firm $Y$. Each firm has marginal cost $c^X$ and $c^Y$ respectively. The firms face a linear demand where $P(Q)=a-b Q$ and the total production of the economy $Q=x+y$. This risk neutral agent maximizes profits so his/her utility will be:

$U(x,y)=w(x(a-b(x+y)-c^X))+(1-w)(y(a-b(x+y)-c^Y))$

If I take first order conditions to maximize this utility I get:

$(a-2b x-c^X)w-by=0$

$(a-2b y-c^Y)(1-w)-bx=0$

Which solves to:

$x=\frac{(1-w)(2 c^X w-c^Y+a(1-2 w))}{b(1-2w)^2}$

$y=\frac{w(2 c^Y(1- w)-c^X-a(1-2 w))}{b(1-2w)^2}$

Assuming this is all correct, I don't understand why when $w=0$, then $y=0$!!! and $x=\frac{a-c^Y}{b}$ maxing the utility $U(x,y)=0$.

I think this doesn't make sense and can't be the optimum because having $x=0$ and $y=\frac{a-c^Y}{2b}$ (monopoly production) would give $U(x,y)=\frac{(a-c)^2
}{2b}>0$

I must have something wrong, the derivatives and solutions are correct, does anyone see what I'm missing here?

**Tags:** production-function, productivity, profit-maximization

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## 2 answers

### Answer by username_1 on 2018-08-09. Score: 1 | Accepted answer

Seems like you are assuming there will be an interior solution.
$$
(a-2b x-c^X)w-by = 0
$$
only needs to hold if the value of $x$ is positive. If it is not, but $x = 0$, the lowest possible amount, then
$$
(a-2b x-c^X)w-by < 0
$$
is not a contradiction. Further decreasing $x$ would increase utility, but it is not possible. Therefore in the case when $w = 0$ and
$$
(a-2b x-c^X)w-by = -by
$$
you do not necessarily have $y = 0$.


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### Answer by username_2 on 2018-08-10. Score: 0

Similar to the corner solution argument by denesp:

When we solve Max U(x,y) when w = 0, we are selecting x and y such that U is maximized. But, w = 0, implies that U is decreasing in x. Thus, we will choose the smallest possible value of x. For any value of y we can select, a non-zero value of x will lower utility.


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