# Thread: Econometrics: MR, Revenue, profit.

1. ## Econometrics: MR, Revenue, profit.

The cost (in 000's) of producing x thousands of loudspeaker systems is TC = x^3 + 3x^2 - 10x + 2. Price (p) and quantity demanded (x) which are both required to be non-negative are related by p = 130 - 2x. Marginal Cost (MC) = 3x^2 + 6x - 10.

a) Find Marginal Revenue (MR) as a function of output
b) What quantity maximises revenue, and what is the corresponding revenue?
c) What quantity maximises profit, and what is the corresponding profit?

2. Hello!
a) $\displaystyle MR = \frac{\mbox{d}TR}{\mbox{d}x} = \frac{\mbox{d}}{\mbox{d}x}(P \cdot x) = \frac{\mbox{d}P}{\mbox{d}x} \cdot x +P \cdot \frac{\mbox{d}x}{\mbox{d}x} = \frac{\mbox{d}P}{\mbox{d}x} \cdot x +P$

Since $\displaystyle P = 130 - 2x$, $\displaystyle \frac{\mbox{d}P}{\mbox{d}x}=-2$ so

$\displaystyle MR = -2x + 130-2x = 130 -4x$.

b) $\displaystyle TR = P \cdot x = (130-2x)x = -2x^2+130x$
To find the quantity that maximizes revenue $\displaystyle TR$, set $\displaystyle \frac{\mbox{d}TR}{\mbox{d}x} = 0$. This gives $\displaystyle -4x+130 = 0$, calculate $\displaystyle x$ and use it to calculate the corresponding revenue $\displaystyle TR$. Check that it really is a maximum.

c) Profit is equal to total revenue $\displaystyle TR$ minus total cost $\displaystyle TC$. To find the maximum profit with respect to quantity sold $\displaystyle x$, set

$\displaystyle \frac{\mbox{d}}{\mbox{d}x}(TR-TC)=0$. You get

$\displaystyle \frac{\mbox{d}}{\mbox{d}x}\left((-2x^2+130x)-(x^3 + 3x^2 - 10x + 2)\right) =$

$\displaystyle = \frac{\mbox{d}}{\mbox{d}x}(-x^3-5x^2+140x-2) = -3x^2-10x+140 =0$, find $\displaystyle x$ (discard negative solution)

and use it to calculate corresponding profit $\displaystyle TR-TC$. Check that it really is a maximum.

Maybe I forgot to multiply by 1000 somewhere but I hope you get the idea.