The average cost is the cost per item of producing a certain number of items, or c(x)/x. So, to find average cost for a certain, given a cost function, simply plug in the number of items into the cost function, and divide by that number. The result is the cost per item produced. The graph of the average cost can be shown on a graphing calculator by inputting the function of the average cost, its minimum, and bounds.
I want to look at a problem that explores average cost just a little bit more. The cost function for your magic broom company is c(x) equals 300,000 plus 60x minus 0.03x² plus 0.000009x³. Now this is in galleons. This is the total cost of producing x magic brooms. A says find the average cost per broom as a function of the number x of brooms produced.
Well, before I get into average cost, this is a graph of the cost function. This function I have here. Average cost remember is a(x) equals c(x), the total cost, divided by x. Now how would this quantity be represented on a graph of average cost? Remember this is y equals c(x).
If I picked an x value here, the average cost would be this value which is c(x) divided by x. The way to represent that is the slope of this line. Now every point on this curve is going to have a different slope. Well, there might be some that have the same, but the idea is that you can calculate the average cost by calculating the slope of this.
So this begs the question, can we find a place where there is a minimum average cost? That's what we're going to do in part b. But for now let's come up with a formula for the average cost. We start with this, and we divide this function by x. I'll just do that termwise. So a(x) is going to be 300,000 over x plus 60x over x is 60, minus 0.03x² over x, minus 0.03x. Then plus 0.000009x³ over x, that will be x². That's my average cost function.
So in part b I'm asked to graph the average cost function on a graphing calculator. I'm asked for what x value is average cost minimum, and what is the minimum average cost. So I'm going to graph this function on my TI-SmartView, it's TI-84. I'm going to graph this, and find the minimum average cost, and where it occurs. So let's take a look at the TI-84.
Here we are looking at the TI-84. I've taken the liberty of already entering in the function a(x). This is my average cost function. Not the cost function. I've set the window to Xmin is 0, Xmax is 600. So this is not the number of brooms. I'm stepping by thousands of brooms. Then in the y-axis, I have the cost in galleons per brooms. So I'm going from 0 to 500 there. You recall in the previous exercise, we had an average cost of about 355 galleons per broom.
Let's graph this. So this is what the graph of average cost looks like. You can see that there is an actual absolute minimum here. We just need to find it. Instead of using calculus to find it, I'm going to use my calculator. If you look at second calc, in the Calculate menu, third thing; minimum. I'll hit number 3. I need a Left Bound.
So I'm going to pick a point that's to the left of the minimum. I'm guessing that 2,000 might be left. So I'm, going to type that in, 2,000 enter. That's good. On right Bound, I'm going to try 5,000. Now that tells the computer where to search for the minimum. It will search between 2,000 and 5,000.
Now it's asking for a guess. I want to just type in 3000. That's between the two. It's found x equals 3,247, that's the number of broom where the average cost is a minimum, and 149.87, that's the number of galleons per broom that is the minimum average cost. So the answer to our problem is that we should produce 3247 brooms to get a minimum average cost. That minimum average cost will be about 150 galleons per broom.
Please enter your name.
Are you sure you want to delete this comment?
Experience the 'A-Ha!' moment with the best teachers
whom we hand-picked for you!
PhD. in Mathematics, University of Rhode Island B.S. in Mechanical Engineering, Cornell University
He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.
“Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”
“I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”
“Hahaha, his examples are the same problems of my math HW!”