1
$\begingroup$

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

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 $y = 0$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.