Differentiation excercise Watch

gr8wizard10
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Question:

A company has a demand function: q=400-2p
and a total cost function: TC=4q^3-116q^2+20q+1500
At what level of output is profit maximised?


What I understand from this is that I have to do TR-TC=TP and then find maximum turning point for TP and the value of q should be the output (quantiy) needed, no?

To fullfill the above aforementioned equation I'll have to do p*q=TR

So given that q is the variable that we have to find

and P=(400-q)/2 [rearranging the demand function]

TR=((400-q/2)*q => ((400q-q^2)/2)

So TP = [((400q-0.5q^2)/2)] - [4q^3-116q^2+20q+1500]

So simlifying this I'm left with : TP = 180q+115.5q^2-4q^3-1500

Differentiating leaves me with => dTP/dQ = 180+231q-12q^2

equating this to 0 and using quadratic formula leaves me with : q=3/4 and q=20

Subbing in the value of 20 to the 2nd derivative: d^2TP/dQ^2 = 231-24q leaves me with -239 which I gather is the maximum turning point.

Am I right in concluding that 20 will be the level of output that profit is maximised where the profit in this case can be calculated by subbing that value in the equation to solve for TP?
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gr8wizard10
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bump please
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ztibor
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(Original post by Anonynous)
Question:

A company has a demand function: q=400-2p
and a total cost function: TC=4q^3-116q^2+20q+1500
At what level of output is profit maximised?


What I understand from this is that I have to do TR-TC=TP and then find maximum turning point for TP and the value of q should be the output (quantiy) needed, no?

To fullfill the above aforementioned equation I'll have to do p*q=TR

So given that q is the variable that we have to find

and P=(400-q)/2 [rearranging the demand function]

TR=((400-q/2)*q => ((400q-q^2)/2)

So TP = [((400q-0.5q^2)/2)] - [4q^3-116q^2+20q+1500]

So simlifying this I'm left with : TP = 180q+115.5q^2-4q^3-1500

Differentiating leaves me with => dTP/dQ = 180+231q-12q^2

equating this to 0 and using quadratic formula leaves me with : q=3/4 and q=20

Subbing in the value of 20 to the 2nd derivative: d^2TP/dQ^2 = 231-24q leaves me with -239 which I gather is the maximum turning point.

Am I right in concluding that 20 will be the level of output that profit is maximised where the profit in this case can be calculated by subbing that value in the equation to solve for TP?
You are right

There is some mistyping in the following rows:

TR=((400-q/2)*q => ((400[/B]q-q^2)/2)

So TP = [((400q-0.5q^2)/2)] - [4q^3-116q^2+20q+1500]

but the TP is OK

That q=20 is the maximum you can examine easily
Substituting q<20 in the dTP/dQ: dTP/dQ(0)=180>0 -> positive -> TP is increasing
Substituting q>20 in the dTP/dQ. dTP/dQ(100)=23280-120000 <0 -> negative -> TP is decreasing
So the turning point of TP at q=20 is maximum
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