Answer to Question #315446 in Calculus for Jack Paul

Question #315446

A firm's average revenue function 


𝐴𝑅=βˆ’18βˆ’7,5𝑄+𝑄

2

.

AR=βˆ’18βˆ’7,5Q+Q2.

Determine the number of units to be produced and sold to maximise revenue.

a.βˆ’1

b.6

c.3,75

d.0


1
Expert's answer
2022-03-23T18:07:46-0400

AR=RQβ‡’R=ARβ‹…Q=(βˆ’18βˆ’7.5Q+Q2)Q==Q3βˆ’7.5Q2βˆ’18QQ3βˆ’7.5Q2βˆ’18Qβ†’max⁑3Q2βˆ’15Qβˆ’18=0β‡’Q∈{βˆ’1,6}AR=\frac{R}{Q}\Rightarrow R=AR\cdot Q=\left( -18-7.5Q+Q^2 \right) Q=\\=Q^3-7.5Q^2-18Q\\Q^3-7.5Q^2-18Q\rightarrow \max \\3Q^2-15Q-18=0\Rightarrow Q\in \left\{ -1,6 \right\}

Since Q is non-negative, we have only Q=6.

But this value is minimum, not maximum:

Rβ€²β€²(6)=(6Qβˆ’15)∣Q=6=36βˆ’15=21>0R''\left( 6 \right) =\left( 6Q-15 \right) |_{Q=6}=36-15=21>0

Actually the value of R tends to infinity at infinite Q.

So the problem is incorrect


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