# Maximization

• Jan 14th 2007, 12:26 PM
413
Maximization
The demand for item A is
P=40 -3.5Q
The production of A entails the following average variable costs:
AVC=1.5Q - 35
Fixed Costs are 24.

a) Calculate the revenue maximizing price of A
Revenue= PQ
Revenue= 40Q-3.5Q^2
Revenue' = 40-7Q
Q=40/7
P=40-3.5(40/7)
P=20

am i correct?

b) Calculate the output level that minimizes the Average total cost.

VC=1.5Q^2-3.5Q
TC=1.5Q^2-3.5Q+24
ATC=1.5Q-3.5+24/Q

am i on the right track so far?
• Jan 17th 2007, 02:57 PM
413
i am having trouble solving this eqauation
ATC=1.5Q-3.5+24/Q
• Jan 18th 2007, 02:12 AM
ticbol
Quote:

Originally Posted by 413
i am having trouble solving this eqauation
ATC=1.5Q-3.5+24/Q

Reason why nobody touches your question is you have variables here that are not defined.
What is P? The demand? Or Price?
What is Q? Quantity? Output?

It looks like your ATC is correct.
Now you want to find Q that will minimize the ATC?

ATC = 1.5Q -3.5 +24/Q
ATC = 1.5Q -3.5 +24Q^(-1)
d(ATC)/dQ = 1.5 +24[(-1)Q^(-2)]
d(ATC)/dQ = 1.5 -24/(Q^2)
Equate that to zero,
0 = 1.5 -24/(Q^2)
Multiply both sides by Q^2,
0 = 1.5Q^2 -24
24 = 1.5Q^2
Q^2 = 24/1.5 = 16
Q = 4
• Jan 18th 2007, 06:38 AM
413
thanks.

I have similar problems that i can't solve, for example this one:
(2/3)Q-12-(20/Q^2)=0

do i do the same thing you did, which i get
(2/3)Q^3-12Q^2-20=0

then what next?
• Jan 18th 2007, 09:51 AM
ticbol
Quote:

Originally Posted by 413
thanks.

I have similar problems that i can't solve, for example this one:
(2/3)Q-12-(20/Q^2)=0

do i do the same thing you did, which i get
(2/3)Q^3-12Q^2-20=0

then what next?

It is the same?
then where is ATC?
Why is (2/3)Q -12 -20/(Q^2) = 0 ?
• Jan 18th 2007, 02:53 PM
413
its a different problem asking for the same thing, i differentiated it and have a problem solving for zero

how do you go about solving for zero for this equation?
(2/3)Q -12 -20/(Q^2) = 0
• Jan 19th 2007, 12:39 AM
ticbol
Quote:

Originally Posted by 413
its a different problem asking for the same thing, i differentiated it and have a problem solving for zero

how do you go about solving for zero for this equation?
(2/3)Q -12 -20/(Q^2) = 0

Again, if it is the same or a similar thing, then why did you get
(2/3)Q -12 -20/(Q^2) = 0 ?

To follow the previous example, should it not be
ATC = (2/3)Q -12 +20/Q ?

--------------------------------
In what you did, after diferrentiating, and setting d(ATC)/dQ to zero, you got that
(2/3)Q -12 -20/(Q^2) = 0
That means the ATC was
ATC = (1/3)Q^2 -12Q +20/Q
Is that so?

-------------------------------------
Okay, whatever, let's say you want to find Q in the equation
(2/3)Q -12 -20/(Q^2) = 0.
Clear the fractions, multiply both sides by 3*Q^2,
2Q^3 -36Q^2 -60 = 0
Q^3 -18Q^2 -30 = 0 ------------(1)
That is a cubic equation.

You can use the Cubic Formula, Cardano's version, if you know how to do that and you have patience.

Or, you can use approximation by iteration. Iteration is repitition. You can do it on your own (method) by substituting values for Q and see if the equation is true, equals zero, or very near zero.
You see that if Q=18, then Q^3 = 18^3 and 18Q^2 = 18^3 also, so you're left with -30 = 0 in the equation. Close.
Then try Q=17.9 or Q=18.1.
Etc.. until you get as close to zero as you like.

Or, if you know the Newton's Method of iteration, then use it. Less guessing here, but that depends on you. [I am fairly fast with my own method. My own method is your own method mentioned above. It is the method of everybody.]

Q1 = Qo -[f(Qo) /f'(Qo)] ----------(i)
where
Qo = seed or initial Q
Q1 = Q closer to the correct root than Qo.

f(Q) = Q^3 -18Q^2 -30 ---------***
f'(Q) = (1/3)Q^2 -36Q ----------***

Say Qo = 18
So, f(18) = 18^3 -18(18^2) -30 = -30
f'(18) = (1/3)(18^2) -36(18) = -540
Q1 = 18 -[(-30)/(-540)] = 17.944

Repeat the procedure.
Q2 = Q1 -[f(Q1)/f'(Q1] ----------(ii)
f(17.944) = (17.944)^3 -18(17.944)^2 -30 = -48.031
f'(17.944) = (1/3)(17.944)^2 -36(17.944) = -538.655
Q2 = 17.944 -[(-48.031)/(-538.655)] = 17.855 --------2nd iteration.

Get to Q3 or Q4, or 3rd or 4th iteration, and you are very close to the correct Q.

Once you get Q here, you might still have 2 more Q's, real number Q's, left, because a cubic equation has exactly 3 roots.
You know how to continue?
That one Q you got from iteration will be a factor (Q -17.---). Divide the original equation, (1), by that factor to get a quadratic equation. Then use the Quadratic Formula there to get the other two roots.

You deal with a cubic equation, so, be patient.