Cost Function C(q) = .05q^3 - 3q^2 + 70q + 80; ave cost and

[kackeret]

Hi everybody-

I'm brand new to this site so hopefully you all can help me!

C(q) = .05q^3 - 3q^2 + 70q + 80 and I found the Ave cost to be: .05q^2 - 3q + 70 + 80/q

a) For what values of q is the ave cost per item increasing? Decreasing? Explain. (Can't be graphically)

b) For what value of q is the ave cost per item smallest? What is the smallest ave cost per item at that point? (Can't be graphically)

Hope you can help!

 

[mmm4444bot]

I'm not sure how far you've gotten on this exercise, so I'm not sure where to start.

Do we need to start by defining the derivative?

Do you know why you're stuck?

Pick a name for the Average Cost function, like A.

A(q) = 0.05q^2 - 3q + 70 + 80/q

The slope of this function's graph is given by the first derivative function A'(q). Instead of an x-axis, we call the horizontal axis the q-axis. The y-axis gets renamed the A-axis.

When A'(q) is zero, then the slope of the graph of the Average Cost function A(q) changes at that very value of q. All of the places where a function's first derivative is zero (or undefined) break up the Real number line into intervals.

Analyze the sign of A'(q) in each of these intervals; those where it's positive are the same intervals where the slope of the Average Cost function is positive, and therefore INCREASING.

The interals where A'(q) is negative are where the Average Cost function is DECREASING.

Again, where A'(q) is zero, the graph of A(q) changes direction, so those values of q are exactly where you need to look for "highest" and "lowest" levels of average cost.

Can you try to find the first derivative A'(q) ?

If you would like more help, then please show whatever work you can accomplish, and try to say something about why you're stuck, so people might determine how to continue helping you.

 

[Deleted member 4993]

Re: Cost Function

Hi everybody-

I'm brand new to this site so hopefully you all can help me!

C(q) = .05q^3 - 3q^2 + 70q + 80 and I found the Ave cost to be: .05q^2 - 3q + 70 + 80/q

a) For what values of q is the ave cost per item increasing? Decreasing? Explain. (Can't be graphically)

b) For what value of q is the ave cost per item smallest? What is the smallest ave cost per item at that point? (Can't be graphically)

Hope you can help!

However, you can check your answer graphically - with your graphic calculator.

 

[kackeret]

Can you try to find the first derivative A'(q) ?

If you would like more help, then please show whatever work you can accomplish, and try to say something about why you're stuck, so people might determine how to continue helping you.

Well I found the a'(q) = .1q - 3 - 80q^-2. So my understanding is that to find where it's increasing and decreasing, I need to set that equal to zero. I guess I'm confused on how to do that. Graphically, I found that the a(q) would be decreasing from 0 to 30.8 and increasing from 30.8 to infinite. (I'm not sure if that's even correct). But I'm getting stuck on finding it algebraically.

Thanks!

 

[Deleted member 4993]

Can you try to find the first derivative A'(q) ?

If you would like more help, then please show whatever work you can accomplish, and try to say something about why you're stuck, so people might determine how to continue helping you.

Well I found the a'(q) = .1q - 3 - 80q^-2. So my understanding is that to find where it's increasing and decreasing, I need to set that equal to zero. I guess I'm confused on how to do that. Graphically, I found that the a(q) would be decreasing from 0 to 30.8 and increasing from 30.8 to infinite. (I'm not sure if that's even correct). But I'm getting stuck on finding it algebraically.

Thanks!

\(\displaystyle 0.1q \, - \, 3 \, - \, \frac {80}{q^2} \, = \, 0\) ....this is how.

However, if you need to solve for 'q' - that would be horse of a different color!!

There are several ways to tackle the problem - graphical is is one of them (which you are not allowed to use).

Other method is numerical approximation - I don't know what you know about those.

I would approximate 'q' (for a'(q) = 0) by noting the fact that for relatively modest 'q' we can ignore "80/q^2" compared to "3" or "0.1*q".

so for relatively large 'q', we would have

0.1*q - 3 = 0

q = 30

then cheque sign of a'(q) at q = 30 , 31 and 32 to decide how does the cost curve [a(q)] behave.

 

[mmm4444bot]

… I found the [first derivative] … my understanding is … to find where it's increasing and decreasing …

In this exercise, we do not care where the derivative is increasing or decreasing.

In other words, the exercise asks for the values of q where A(q) is increasing/decreasing, not A'(q).

Maybe you're thinking correctly, but somehow did not type what you're actually thinking.

'

… I need to set that equal to zero. I guess I'm confused on how to do that …

Again, you don't seem to be typing what you're thinking. As Subhotosh noted, it's easy to set something equal to zero. You just do it.

Maybe, you are actually thinking the following.

"I guess I'm confused on how to find the values of q that make A'(q) equal to zero."

Is finding the zeros of A'(q) what you're trying to comment about?

(I think so. It appears that Subhotosh thinks so, too.)

'

… Graphically, I found that the a(q) would be decreasing from 0 to 30.8 and increasing from 30.8 to [infinity]. (I'm not sure if that's even correct). …

Why are you doubting your graph? Did you draw the graph of A(q) by hand, after calculating data points, and you're not confident that you did this correctly?

If, instead, you let some machine plot the graph for you, followed by zooming in on the local minimum to find the approximation for q, then you need only verify that you properly entered the definition of function A.

In other words, I'm not sure why you're doubting the value q = 30.8 because it should be clear from the graph. Anyway, 30.8 is the correct zero of A'(q), rounded to one decimal place.

(Maybe you're not actually doubting the value of 30.8. Maybe you're doubting something else, and you did not type what you're actually thinking.)

At this point, I need to know what instructions you've been given with respect to technology. You stated that you're stuck on finding the value 30.8 algebraically. Are you supposed to do this?

If your instructor integrates technology as a tool in this class, then it could be that your instructor expects students to let some machine find the zero of A'(q) in this exercise.

If this is the case, then forget about trying to solve A'(q) = 0 algebraically; let your machine do the calculation for you.

I recently spent 1 hour 15 minutes with a student at college who needed to solve a cubic equation. We worked through it by hand and obtained the solution. A few days later, I discovered that he was expected to use a graphing calculator to solve his word problem. Because the student did not communicate clearly, that 1 hour 15 minutes is a waste of his time and mine.

Do you have any more questions about how to complete your exercise? If you do, then please clue us in to how technology is used in your class, and try to be more careful about expressing your thoughts.

 

[kackeret]

Hi-

I really appreciate all of everybody's help. I had to turn the work in today. Your tips will be helpful for communicating my thoughts and the problems more clearly in the future. This is the first time I have used this site and I'm really greatful for everybody's help!

Thanks again, Kyle

 

[mmm4444bot]

I had to turn the work in today …

Thank you for clarifying. Here are the answers.

Part (a): The average cost per item is decreasing when q is less than 30.8 because the sign of the derivative is negative within this interval. The average cost per item is increasing when q is greater than 30.8 because the sign of the derivative is positive within this interval.

Part (b): The smallest average cost per item is 27.63, and this value is achieved when q = 30.8.

Of course, your work shown also contributes to the requested explanations.

 

[BigGlenntheHeavy]

The Cost Function

First, a few particulars to alleviate all and any angst, real or imagine.

C(q) = .05q^3-3q^2+70q+80, cost function, but what does it mean?

q is the number of units produced and C(q) is the cost of producing those units; for example to produce one unit it would cost C(1) = $147.50, to produced i00 units, it would cost C(100) = $27,080.00.

Hence, assuming we aren't producing 35.8973456 units, we have our domain, q is greater or equal to one, q a counting number. Also assuming that it cost something to produce at least one unit (we live in a democracy, people won't work for nothing, plus the raw material cost, we have our range, C(q) is greater than zero.

However this problem revolves around the average cost of producing one unit, so lets call the average cost A(q) as already mention. Hence, we will be working with A(q) = .05q^2-3q+70+80/q.

Now A' (q) = .1q-3-80/q^2. Set the slope to zero gives q = 30.8411. Now since we can't have 30.8411 units. we will round off to 31 units.

A(31) = $27.63, This means to produce 31 units, the cost is $27.63 per unit. A(30) = $27.67, A(32) = $27.70, hence A(31) must be a relative min (critical number). Another check (2nd deravitive). A"(q) = .1+160/q^3, A"(31) = .105371 > 0, therefore A(31) is a relative if not absolute minimum.

Now the limit as q goes to one from the positive side, C(q) goes to 147.05 and as q goes to infinity, C(q) goes to infinity.
Ergo we have a graph which is concave up with A(31) = $27.63 as our only critical point (absoute minumum). Domain [1,?),Range [27.63, ?).

Therefore the Average cost function is decreasing from 1 to 31 units and increasing from 32 to infinity units: also 31 units produced will give us the lowest cost per unit, namely $27.63 per unit.

 

[Deleted member 4993]

The Cost Function


.....Hence, assuming we aren't producing 35.8973456 units, we have our domain, q is greater or equal to one, q a counting number. The unit could be a continuous variable, like weight or length of material produced.

Also assuming that it cost something to produce at least one unit (we live in a democracy, people won't work for nothing, plus the raw material cost, we have our range, C(q) is greater than zero......

 

[mmm4444bot]

… q is the number of units produced and C(q) is the cost of producing those units [in dollars] …

I think that these two assumptions are reasonable, but, after previously considering them, I chose to not make them. This is why I did not round 30.8 and why I did not put a dollar sign in front of the number 27.63.

Sometimes, this issue matters. For example, it would not make sense to tell somebody that you have 3.2 children.

Sometimes, this issue does not matter. For example, it makes sense to report some average number of children per family as 3.2.

This issue is similar to the issue of rules regarding significant figures. Sometimes, these rules matter; with most introductory-course word problems, they do not.

And, as Subhotosh noted, we don't really know what is being produced in terms of what.

(I defer this issue to the original poster's instructor, heh, heh.)