To identify the profit, there is a need to set the equation: MR = MC
Given:
Q = 144 P 2 , P 2 = 144 Q Q = \frac{144}{P^2}, \space P^2 = \frac{144}{Q} Q = P 2 144 , P 2 = Q 144
Or P = ( 144 Q ) 1 2 . . . . . . . . . . . ( i ) P=(\frac{144}{Q})^{\frac{1}{2}} ...........(i) P = ( Q 144 ) 2 1 ........... ( i )
Therefore, P = ( 12 Q ) P=(\frac{12}{\sqrt{Q}}) P = ( Q 12 )
Now TR = Price and Quantity sold
TR = PQ
= 12 Q × Q =\frac{12}{\sqrt Q}\times Q = Q 12 × Q
= 12 Q . . . . . . . . . . . . . . . . . . . . . . . . . . ( i i ) =12\sqrt Q..........................(ii) = 12 Q .......................... ( ii )
Now, MR=
Δ ( T R ) Δ Q \frac{\Delta(TR)}{\Delta Q} Δ Q Δ ( TR ) [MR derivative of T R Δ Q \frac{TR}{\Delta Q} Δ Q TR ]
= Δ 12 Q Δ Q =\frac{\Delta 12\sqrt Q}{\Delta Q} = Δ Q Δ12 Q
= 1 2 ( 12 Q ) = 6 Δ Q . . . . . . . . . . . . . . . . . . . . . . . . . ( i i i ) =\frac{1}{2}(12Q)\\=\frac{6}{\Delta Q}.........................(iii) = 2 1 ( 12 Q ) = Δ Q 6 ......................... ( iii )
Given the total fixed cost of $5
Average variable cost (AVC)= Q 1 2 = Q =Q^{\frac{1}{2}}=\sqrt Q = Q 2 1 = Q
TC = TFC + TVC
TVC = (AVC)(Q)
Therefore, total cost T C = 5 + ( Q ) ( Q ) TC=5+(\sqrt Q)(Q) TC = 5 + ( Q ) ( Q )
= 5 + Q 1 2 + 1 = 5 + Q 3 2 . . . . . . . . . . . . . . . . . . . . . . . ( i v ) =5+Q^{\frac{1}{2}+1}\\=5+Q^{\frac{3}{2}}.......................(iv) = 5 + Q 2 1 + 1 = 5 + Q 2 3 ....................... ( i v )
with TC, Now we can compute MC by taking the derivation within respect to Q
Now MR = MC,
Therefore to find the profit-maximizing level of output
6 Q = 3 2 Q o r Q = 12 3 \frac{6}{\sqrt Q}=\frac{3}{2}\sqrt Q\\or\space Q=\frac{12}{3} Q 6 = 2 3 Q or Q = 3 12
Q = 4units
by putting Q = 4
P = 12 Q P=\frac{12}{\sqrt Q} P = Q 12
P = 12 4 P=\frac{12}{\sqrt 4} P = 4 12
P = 12 2 P=\frac{12}{2} P = 2 12
P = $6
Then profit is:
Profit = PQ - TC
= ( 6 ) ( 4 ) − [ 5 + ( 4 ) 3 2 ] =(6)(4) - [5 + (4)^{3}{2}] = ( 6 ) ( 4 ) − [ 5 + ( 4 ) 3 2 ]
= 24 − 13 = $ 11 =24-13\\
= \$11 = 24 − 13 = $11
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