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A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
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gmatt1476 wrote:

A certain manufacturer uses the function \(C(x) = 0.04x^2 – 8.5x + 25,000\) to calculate the cost, in dollars, of producing x thousand units of its product. The table above gives values of this cost function for values of x between 0 and 50 in increments of 10. For which of the following intervals is the average rate of decrease in cost less than the average rate of decrease in cost for each of the other intervals?

A. From x = 0 to x = 10
B. From x = 10 to x = 20
C. From x = 20 to x = 30
D. From x = 30 to x = 40
E. From x = 40 to x = 50

PS18871.01

Attachment:
2019-09-21_1558.png



Hi RK007,
The average rate of decrease in cost --This means decrease in cost per unit and as the number of units is same in each, that is 10,000 units, we have to divide the decrease in cost by 10,000.

Now, Since the change in cost will be \(\frac{previous - this}{10,000}\) . We are looking for the least value of \(\frac{previous - this}{10,000}\) and the denominator is the SAME so it will depend on numerator( previous - this)..
So basically the answer will be the minimum drop from the previous, so it will
0-10........81
10-20......73
20-30....65
30-40....57
40-50.....41...Our answer

Another way..
We are looking at a quadratic equation \(C(x)=0.04x^2-8.5x+25000\), and the coefficient of x^2 is positive, 0.04, so it will give us a U parabola, which becomes less steeper as we move up..
Hence more the value of x, lesser will be the decrease. Here 40-50 is the max interval given, so the average rate of decrease in cost will be least here..

E

However, a bit disappointed with the wordings on an official question..
A certain manufacturer uses the function \(C(x) = 0.04x^2 – 8.5x + 25,000\) to calculate the cost, in dollars, of producing x thousand units of its product.
This tells us that against 50, the cost of 50,000 units is given by 24,675 and 40,000 units is given by 24724...
But, ofcourse the question is meant to tell us that as the number of the items produced increases, the cost per 10,000 unit decreases and that is given by the table..
so what we have C(X) is the cost of 10,000 items when the production is increased to x thousand items..
General Discussion
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A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
gmatt1476 wrote:

A certain manufacturer uses the function \(C(x) = 0.04x^2 – 8.5x + 25,000\) to calculate the cost, in dollars, of producing x thousand units of its product. The table above gives values of this cost function for values of x between 0 and 50 in increments of 10. For which of the following intervals is the average rate of decrease in cost less than the average rate of decrease in cost for each of the other intervals?

A. From x = 0 to x = 10
B. From x = 10 to x = 20
C. From x = 20 to x = 30
D. From x = 30 to x = 40
E. From x = 40 to x = 50

PS18871.01

Attachment:
2019-09-21_1558.png


Hello Bunuel chetan2u other experts,
Could you please help me with this problem?
What is the exact meaning of 'the average rate of decrease in cost'?
And the meaning of 'the average rate of decrease in cost for each of the other intervals'?

Thanks in advacne :)

Originally posted by RK007 on 13 Oct 2019, 01:42.
Last edited by RK007 on 17 Oct 2019, 07:29, edited 1 time in total.
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Re: A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
Hello VeritasKarishma

Can you help with the above question? Thank you
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Re: A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
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Let x be lower value of the interval and y be the higher (given diff of interval =10) then
Avg decease in cost= [C(y)-C(x)]/10
Solving the equation
= [0.04(y^2-x^2)-8.5(y-x)]/10 but y-x=10
so Avg= [0.04(y^2-x^2)-85]/10
For Avg will minimum if y^2-x^2 is maximum that will only happen for interval 40-50
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Re: A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
chetan2u wrote:
gmatt1476 wrote:

A certain manufacturer uses the function \(C(x) = 0.04x^2 – 8.5x + 25,000\) to calculate the cost, in dollars, of producing x thousand units of its product. The table above gives values of this cost function for values of x between 0 and 50 in increments of 10. For which of the following intervals is the average rate of decrease in cost less than the average rate of decrease in cost for each of the other intervals?

A. From x = 0 to x = 10
B. From x = 10 to x = 20
C. From x = 20 to x = 30
D. From x = 30 to x = 40
E. From x = 40 to x = 50

PS18871.01

Attachment:
2019-09-21_1558.png



Hi Chetan

Hi RK007,
The average rate of decrease in cost --This means decrease in cost per unit and as the number of units is same in each, that is 10,000 units, we have to divide the decrease in cost by 10,000.

Now, Since the change in cost will be \(\frac{previous - this}{10,000}\) . We are looking for the least value of \(\frac{previous - this}{10,000}\) and the denominator is the SAME so it will depend on numerator( previous - this)..
So basically the answer will be the minimum drop from the previous, so it will
0-10........81
10-20......73
20-30....65
30-40....57
40-50.....41...Our answer

Another way..
We are looking at a quadratic equation \(C(x)=0.04x^2-8.5x+25000\), and the coefficient of x^2 is positive, 0.04, so it will give us a U parabola, which becomes less steeper as we move up..
Hence more the value of x, lesser will be the decrease. Here 40-50 is the max interval given, so the average rate of decrease in cost will be least here..

E

However, a bit disappointed with the wordings on an official question..
A certain manufacturer uses the function \(C(x) = 0.04x^2 – 8.5x + 25,000\) to calculate the cost, in dollars, of producing x thousand units of its product.
This tells us that against 50, the cost of 50,000 units is given by 24,675 and 40,000 units is given by 24724...
But, ofcourse the question is meant to tell us that as the number of the items produced increases, the cost per 10,000 unit decreases and that is given by the table..
so what we have C(X) is the cost of 10,000 items when the production is increased to x thousand items..



Hi chetan2u

How can value of Y decrease with increasing X in a U SHAPED parabola. Can you please explain.
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A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
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chetan2u wrote:
The average rate of decrease in cost --This means decrease in cost per unit and as the number of units is same in each, that is 10,000 units, we have to divide the decrease in cost by 10,000.

Now, Since the change in cost will be \(\frac{previous - this}{10,000}\) . We are looking for the least value of \(\frac{previous - this}{10,000}\) and the denominator is the SAME so it will depend on numerator( previous - this)..
So basically the answer will be the minimum drop from the previous, so it will
0-10........81
10-20......73
20-30....65
30-40....57
40-50.....41...Our answer

Dear IanStewart,

I'm not sure what The average RATE of decrease in cost means.
Why isn't The average RATE of decrease in cost = \(\frac{AVG. previous - AVG. this}{AVG. PREVIOUS}\)?

In other words, I thought the equation would be [C(a+10)/(a+10) - C(a)/a]/C(a+10)/(a+10)
Aren't we looking for a percentage decrease?

Originally posted by kornn on 20 Jul 2020, 23:20.
Last edited by kornn on 20 Jul 2020, 23:52, edited 5 times in total.
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A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
chetan2u wrote:
Another way..
We are looking at a quadratic equation \(C(x)=0.04x^2-8.5x+25000\), and the coefficient of x^2 is positive, 0.04, so it will give us a U parabola, which becomes less steeper as we move up..
Hence more the value of x, lesser will be the decrease. Here 40-50 is the max interval given, so the average rate of decrease in cost will be least here..

Dear IanStewart,

I have similar doubt as zs2 had.
Please confirm whether the highlighted part is incorrect.
Chetan was right that the equation is U-shaped parabola.
However, the more value of x, the GREATER will be the INCREASE of C(x) (as attached herein).

I think he is confused that the question is actually asking for C(x) DIVIDED BY x, not just C(x) itself.
Attachments

graph.PNG
graph.PNG [ 49 KiB | Viewed 20470 times ]

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Re: A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
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varotkorn wrote:
chetan2u wrote:
Another way..
We are looking at a quadratic equation \(C(x)=0.04x^2-8.5x+25000\), and the coefficient of x^2 is positive, 0.04, so it will give us a U parabola, which becomes less steeper as we move up..
Hence more the value of x, lesser will be the decrease. Here 40-50 is the max interval given, so the average rate of decrease in cost will be least here..

Dear IanStewart,

I have similar doubt as zs2 had.
Please confirm whether the highlighted part is incorrect.
Chetan was right that the equation is U-shaped parabola.
However, the more value of x, the GREATER will be the INCREASE of C(x) (as attached herein).

I think he is confused that the question is actually asking for C(x) DIVIDED BY x, not just C(x) itself.



The effect on C(x) will be the same as effect on C(x)/x or C(x)/100 or C(x)/y till the time x and y are positive integers.
So if C(x) is decreasing in foll0wing manner 1000, 900, 800 and x is 10, then the decrease in C(x)/x will be 100, 90, 80.....
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varotkorn wrote:
I'm not sure what The average RATE of decrease in cost means.
Why isn't The average RATE of decrease in cost = \(\frac{AVG. previous - AVG. this}{AVG. PREVIOUS}\)?

In other words, I thought the equation would be [C(a+10)/(a+10) - C(a)/a]/C(a+10)/(a+10)
Aren't we looking for a percentage decrease?


I agree with what Chetan has said above - there is something wrong with the wording of the question. I've seen three official Quant questions in my life that were flawed for one reason or another (out of about ten thousand I've seen in total, so it's extremely rare) but this is one of them. If you read the question literally, the chart is saying that the company will pay $25,000 in order to produce zero products, but will pay less than that in total if they instead produce 50,000 products. That doesn't make any logical sense -- when you make more things, your cost per unit normally goes down, but your total costs obviously don't go down. But that's what this question literally says. The costs in the C(x) column, as Chetan points out, are surely meant to be per 10,000 units (or something like that). Then it would make sense that the company saves money per unit as they scale up their production.

This question is really trying to test a concept from calculus, but in a simple enough way that you don't need to know calculus to answer it. When we talk about the rate of decrease or rate of increase of a curve at a point, we're talking about the curve's slope at that point. In calculus you learn how to compute the exact slope of a curve at any point. But if you can approximate the curve using short straight lines, you can approximate the slope of the curve at various points by finding the slopes of those lines. That's what we can do using the values in the table provided in this question -- the table is letting us approximate this curve (which is a parabola) with four line segments. The slope of each line segment tells us how quickly the cost is decreasing. So we want to calculate "rise/run" between each successive pair of points, but for each, "run" is just 10, so we can just compare the change in C(x), the righthand column of the table, and find when the decrease is smallest. That's why the solutions above just using subtraction are perfectly correct.

If you did know calculus (which you don't need on the GMAT), you could also just compute the derivative of C(x), which is 0.08x - 8.5. That's the slope of the curve C(x) at any point x. The slope clearly increases as x does, and we'll get the smaller rate of decrease the larger x is (when the curve is, in fact, decreasing -- it will start increasing when x > 106.25).

varotkorn wrote:
Please confirm whether the highlighted part is incorrect.
Chetan was right that the equation is U-shaped parabola.
However, the more value of x, the GREATER will be the INCREASE of C(x) (as attached herein).


No, Chetan was right. The vertex of the parabola is at x = 106.25. So for the values of x the question is asking about, the parabola is falling. It only starts rising when x becomes larger than 106.25. It falls less and less quickly as you move towards x = 106.25 from the left.

Overall, this is a strange question, and the wording is problematic. I very much doubt you'll see something similar on your test, so I wouldn't worry about it much.
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A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
Dear IanStewart,

I agree with you that this problem is somewhat flawed.
However, I have some questions in order to arrive at the takeaways for this official question:

Q1. When we see the word RATE (as in the average rate of decrease in cost), will it always mean finding the SLOPE?
At first, I thought that the question asks us to find percentage decrease -- that is, (NEW cost per unit - PREVIOUS cost per unit)/PREVIOUS cost per unit.

Q2. I'm not sure what is the unit for the slope. Is it dollars per unit per unit? I can't make much sense of what the slope actually means.

Q3. Regarding the word the average rate of decrease in cost, what's the purpose of the word AVERAGE?
According to the solution, we aren't finding average -- that is, sum/total number of items.
We're just finding the slope -- that is rise/run.
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kornn wrote:
Q1. When we see the word RATE (as in the average rate of decrease in cost), will it always mean finding the SLOPE?
At first, I thought that the question asks us to find percentage decrease -- that is, (NEW cost per unit - PREVIOUS cost per unit)/PREVIOUS cost per unit.

Q2. I'm not sure what is the unit for the slope. Is it dollars per unit per unit? I can't make much sense of what the slope actually means.

Q3. Regarding the word the average rate of decrease in cost, what's the purpose of the word AVERAGE?
According to the solution, we aren't finding average -- that is, sum/total number of items.
We're just finding the slope -- that is rise/run.


Q1 - Technically the answer is 'yes', but whether you should think of a rate as a 'slope' depends on what you're doing. If you're doing calculus, especially in an applied setting (physics, financial math, etc), then it is a good idea to think of rates that way. You'll do that all the time in an MBA program. If you're doing the GMAT though, it's almost never useful to think of a rate as a slope. That's because rates are almost always constant on the GMAT, and it's only when rates change over time that you start to need the machinery of calculus (which is all about slopes and areas) to really get anywhere, and if you want to understand calculus well, you need to know what a slope represents. It's true that if a car travels at 100 mph, then after 1 hour the car has traveled 100 miles, and after 2 hours it has traveled 200 miles, and if you plot a distance-versus-time graph with the points (1, 100), (2, 200), etc, you get a line, and the slope of that line, 100, is always going to be the speed or 'rate' at which the car is traveling. But I don't see any good reason to think about rates that way if you're doing GMAT questions. This one question is an exception, because the rate changes as x changes (because the function is a curve, not a line) which is why this question is strange, and why I needed to borrow from my calculus knowledge to guess what it was asking.

Q2 - I'd be guessing because of the problems with the wording of the question, but It is something like the change in "cost per unit" per unit produced. So you have the right idea.

Q3 - The rate is constantly changing. The decrease as x goes from 0 to 1 is different from the decrease as x goes from 9 to 10. So the word "average" needs to be there, because it's not a constant value. It's the same idea as with average speed - if a car travels from A to B at one speed, then from B to C at another, you can talk about the car's average speed from A to C, but not simply its "speed" from A to C, because its speed is not constant.
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A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
Dear IanStewart,

Thank you for your crystal clear explanation sir!
IanStewart wrote:
Q3 - The rate is constantly changing. The decrease as x goes from 0 to 1 is different from the decrease as x goes from 9 to 10. So the word "average" needs to be there, because it's not a constant value. It's the same idea as with average speed - if a car travels from A to B at one speed, then from B to C at another, you can talk about the car's average speed from A to C, but not simply its "speed" from A to C, because its speed is not constant.


As for Q3, I'm not sure whether I have a correct understanding:

According to your previous reply: If you did know calculus (which you don't need on the GMAT), you could also just compute the derivative of C(x), which is 0.08x - 8.5. That's the slope of the curve C(x) at any point x.

Slopes at any values of x are continuously changing. Hence, the rate of decrease in cost for each interval (e.g. from x = 0 to x = 10 or from x = 10 to x = 20...) is thus called average.
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Re: A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
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I'm not precisely sure what you're asking, but in calculus (which you don't need on the GMAT), then from one function like the C(x) in this question, you can find another function called the 'derivative'. If you plug a value of x into the derivative, you learn the slope of the original function at that x-value. Technically, you're really finding the slope of a tangent line to the curve at that point. So in calculus, you're dealing with instantaneous things, the rate of change at a single point, what the slope would look like if you zoomed in infinitely close to that point on a curve. In this question, you're not dealing with anything like that -- you're looking at intervals 10 units long, which is why they need to use the word 'average'.
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Re: A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
Dear IanStewart,

I have one last question. (I'm really sorry if I ask too much :please: )
IanStewart wrote:
The vertex of the parabola is at x = 106.25.

I've found in GMAT Club Math book that the vertex of the parabola is -b/2a.
I know that the vertex of the parabola is derived from taking first order derivative and setting the equation to 0 (at the vertex, the slope = 0)

However, I've also noticed that the vertex of the parabola is the same as the root of a quadratic formula (as attached herein) when the term in the radical is equal to 0.

Is it a coincidence?
Attachments

quadratic formula.PNG
quadratic formula.PNG [ 156.04 KiB | Viewed 19822 times ]

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kornn wrote:
Dear IanStewart,

I have one last question. (I'm really sorry if I ask too much :please: )

I've found in GMAT Club Math book that the vertex of the parabola is -b/2a.
I know that the vertex of the parabola is derived from taking first order derivative and setting the equation to 0 (at the vertex, the slope = 0)

However, I've also noticed that the vertex of the parabola is the same as the root of a quadratic formula (as attached herein) when the term in the radical is equal to 0.

Is it a coincidence?


Interesting observation -- no, it's not a coincidence. By symmetry, the vertex of a parabola will be exactly halfway between its x-intercepts. So the quadratic formula automatically needs to produce two solutions (when a quadratic has two solutions) that look like this;

x = vertex + d
x = vertex - d

and that's exactly what the quadratic formula gives us (where d = [√(b^2 - 4ac)]/2a).

You don't need any calculus to find the vertex of a parabola, incidentally. If you have the equation

y = ax^2 + bx + c

changing the number c will just move the parabola up or down, not left or right. So the x-coordinate of the vertex of y = ax^2 + bx + c is the same as the x-coordinate of the vertex of y = ax^2 + bx. By symmetry, the vertex of this parabola will be halfway between its two x-intercepts. If you set y=0, you find the two x-intercepts are at x = 0 and x = -b/a. The vertex is halfway between those, so is the average of those two values, and must be at x = -b/2a.

I've never needed to know anything about how to find the vertex of a parabola to answer an official GMAT question, though.
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Re: A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
Bunuel Could you help with explaining this if possible?
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Re: A certain manufacturer uses the function C(x) = 0.04x^2 – 8.5x + 25,00 [#permalink]
For those who know derivation..

(C(x) = 0.04x^2 – 8.5x + 25,000\)

d(C(x)) = 0.08x-8.5
--------
dx

minimum change in value for equal intervals come around zero slope region.

so, for slope to be 0


d(C(x)) = 0.08x-8.5 =0
--------
dx

hence, 0.08x= 8.5

so, x=106.25

In all options, x=50 is closest to this value
Hence E is the correcct answer.
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