Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A producer produces good y using a single input x according to the production fu

ID: 1181015 • Letter: A

Question

A producer produces good y using a single input x according to the production function y=x^a where 0<a<1. The producer can sell as much as he wants at a unit price p, but must pay wage w for every unit of x used.

1) What are the first and second-order condition for the profit maximization problem?

2) Derive the expression for the profit maximizing level of output y*(p,w,a), the profit maximizing level of input x*(p,w,a), and the optimal profit pi*(p,w,a) = py*(p,w,a)

A producer produces good y using a single input x according to the production function y=x^a where 0

Explanation / Answer

Given: Output, y = x^a


=> Revenue = p*y = p*x^a

=> Cost = w*x


=> Profit = Revenue - Cost = p*x^a - wx ----- (i)


To maximize profit, d(Profit)/dx = 0 and the second order derivative is negative.

Hence, pax^(a-1) -w =0 and the second order derivative = pa(a-1)x^(a-2) <0. Here, p>0, a>0, a-1 <0. So, the second order differential is negative anyway.


1) So, the first and second order condition for the profit maximization problem are

pax^(a-1) -w =0, and

pa(a-1)x^(a-2) <0


2) From the condition, x* = (w/pa)^[1/(a-1)]

Substituting this in equation for y, we get y* = (w/pa)^[a/(a-1)]

From (i), Profit = p * (w/pa)^[(a/(a-1)] - w * (w/pa)^[1/(a-1)]. You can simplify this.


3) When a = 1, y=x. Profit function = py - wx = px - wx = (p-w) x. The first order differential d(profit)/dx = p-w. This is not a function in x. So, the profit is a linear function and the profit is maximum with the maximum input used and maximum output produced.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote