is profit maximisation seen as local minimum or maximum? Is part (b) the correct way of doing first derivative? There is image of the questions, another image of the answers of part (a) and (b). The function is given as,for stationary points,or,dor, (-4Q³+48Q²-117Q-100) = 0dQ-12Q² +96Q-117or,So,Q==-96+√3600-24=-96+60-243624= 1.5df(Q)dQOR Q=f(Q)=-4Q³ +48Q²-117Q-100=0OR Q=QOR Q == 0==-96+√96²-4(-12)(–117)2(-12)15624OR Q = 6.5-96 ± √3600-24-96-√3600-24-96-60-24Hence, the stationary points are 1.5 and 6.5In order the find Q that maximises profit:We can use the table as followsQProfitf(Q)=-4Q³ +48Q²-117Q-100-4(1.5)³ +48(1.5)²-117(1.5)-1001.5-1816.5-4(6.5)³ +48(6.5)²-117(6.5)-10069Thus, using first derivatives test, we find the stationary points and then we put the points in theequation given in order to see at which point does Q maximises profit.We see that, when Q = 6.5, the profit is maximum, so the firm would be better off if it willproduce Q = 6.5 The profit of a firm is described by the following function: f(Q) = -4Q^3 + 48Q^2 -117Q 100 where Q is the quantity produceda)Find the stationary points for this functionb) Using the first derivative test, find for what value of Q is the profit maximised.c) Find the feasible values of Q where the graph of the profit function is concaveup. Show working

Profit maximization refers to the goal of a firm to maximize its net profit, which is the difference…

Answered: s profit maximisation seen as local…

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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is profit maximisation seen as local minimum or maximum? Is part (b) the correct way of doing first derivative? There is image of the questions, another image of the answers of part (a) and (b).

The function is given as,
for stationary points,
or,
d
or, (-4Q³+48Q²-117Q-100) = 0
dQ
-12Q² +96Q-117
or,
So,
Q=
=
-96+√3600
-24
=
-96+60
-24
36
24
= 1.5
df(Q)
dQ
OR Q=
f(Q)=-4Q³ +48Q²-117Q-100
=0
OR Q=
Q
OR Q =
= 0
=
=
-96+√96²-4(-12)(–117)
2(-12)
156
24
OR Q = 6.5
-96 ± √3600
-24
-96-√3600
-24
-96-60
-24
Hence, the stationary points are 1.5 and 6.5
In order the find Q that maximises profit:
We can use the table as follows
Q
Profit
f(Q)=-4Q³ +48Q²-117Q-100
-4(1.5)³ +48(1.5)²-117(1.5)-100
1.5
-181
6.5
-4(6.5)³ +48(6.5)²-117(6.5)-100
69
Thus, using first derivatives test, we find the stationary points and then we put the points in the
equation given in order to see at which point does Q maximises profit.
We see that, when Q = 6.5, the profit is maximum, so the firm would be better off if it will
produce Q = 6.5

The profit of a firm is described by the following function: f(Q) = -4Q^3 + 48Q^2 -
117Q 100 where Q is the quantity produced
a)
Find the stationary points for this function
b) Using the first derivative test, find for what value of Q is the profit maximised.
c) Find the feasible values of Q where the graph of the profit function is concave
up. Show working

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How is my answer correct? from what I heard f < 0 = maximum and f > 0 is minimum. the Q = 6.5 is above 0 which means its minimum and when substituting Q=6.5 to the profit function it gives 69, which above 0 which is minimum, how is it maximum?

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