idempotency
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Hi all,
The problem asks to prove:
e_{Q,P} = \frac{AR}{AR-MR}
In which AR is average revenue and MR is marginal revenue.
Then verify this for demand equation p = a-bx
I developed several steps:
<br /> \begin{flalign*}AR = \frac{TR(Q)}{Q} ; MR = \frac{dTR(Q)}{dQ} \\*<br /> <br /> \frac{AR}{AR-MR} = 1 - \frac{AR}{MR} = 1 - \frac{TR(Q)}{Q} . \frac{dQ}{dTR(Q)}\end{flalign*} (??)
Then I also found that this proof brings me somewhat closer to e(Q,P):
<br /> \begin{flalign*} Q=a-bP\\*<br /> Thus: ~ TR = (a-bP).P = aP-bP^2 (1)\\*<br /> Taking~the~derivatives:<br /> \frac{\partial TR}{\partial P} = a-2bP (2)\\*<br /> <br /> From~(1): P = \frac{a-Q}{b} \\*<br /> Thus~ (a),(b):\\*<br /> \frac{TR}{\partial P} = a+ \frac{a-Q}{b} = 3a-2Q\\*<br /> <br /> Elastic~function: E(P) = \frac{\partial Q}{\partial P} . \frac{P}{Q} \\*<br /> = b . \frac{a-Q}{Qb} (from~(b))\\*<br /> We~have: Q(1+E) = Q(1=\frac{Q-a}{Q} = ... = a + b(2-a)\\*<br /> <br /> <br /> \end{flalign*}
I am a bit stuck here - I am attempting to prove that ∂TR/∂P=Q(1+E) is true (which it is I believe and may go from there.
Am I overcomplicating this? Can you give some hints?
Thanks.
Omaron
Note: Apology for the weird indentation - still trying to figure out LaTeX
The problem asks to prove:
e_{Q,P} = \frac{AR}{AR-MR}
In which AR is average revenue and MR is marginal revenue.
Then verify this for demand equation p = a-bx
I developed several steps:
<br /> \begin{flalign*}AR = \frac{TR(Q)}{Q} ; MR = \frac{dTR(Q)}{dQ} \\*<br /> <br /> \frac{AR}{AR-MR} = 1 - \frac{AR}{MR} = 1 - \frac{TR(Q)}{Q} . \frac{dQ}{dTR(Q)}\end{flalign*} (??)
Then I also found that this proof brings me somewhat closer to e(Q,P):
<br /> \begin{flalign*} Q=a-bP\\*<br /> Thus: ~ TR = (a-bP).P = aP-bP^2 (1)\\*<br /> Taking~the~derivatives:<br /> \frac{\partial TR}{\partial P} = a-2bP (2)\\*<br /> <br /> From~(1): P = \frac{a-Q}{b} \\*<br /> Thus~ (a),(b):\\*<br /> \frac{TR}{\partial P} = a+ \frac{a-Q}{b} = 3a-2Q\\*<br /> <br /> Elastic~function: E(P) = \frac{\partial Q}{\partial P} . \frac{P}{Q} \\*<br /> = b . \frac{a-Q}{Qb} (from~(b))\\*<br /> We~have: Q(1+E) = Q(1=\frac{Q-a}{Q} = ... = a + b(2-a)\\*<br /> <br /> <br /> \end{flalign*}
I am a bit stuck here - I am attempting to prove that ∂TR/∂P=Q(1+E) is true (which it is I believe and may go from there.
Am I overcomplicating this? Can you give some hints?
Thanks.
Omaron
Note: Apology for the weird indentation - still trying to figure out LaTeX