Econometrics: MR, Revenue, profit.

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

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

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

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

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

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

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

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

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

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