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Calculus Thread
TL DR
Not at all confident in his reflexive opinions of thingsRegistered User regular
Not at all confident in his reflexive opinions of thingsRegistered User regular
Hi all, I'm reviewing for an upcoming exam and so I'll be asking for clarification on a few things throughout the evening.
Solved problems in spoilers.
4. A demand function is p = 400 - 2q, where q is the quantity of good sold for price $p. Which of the following is an expression for the total revenue, R, in terms of q?
I just too R = p*q and substituted the function for p to get R = (400-2q)*q Right?
5. The figure below* shows the graphs of marginal cost and marginal revenue. What production level could maximize the profit?
*The graph is a concave-up parabola labeled $/unit (MC) that intersects with a horizontal line labeled MR at x=1000 and x=3000
I hope that's clear, otherwise I can rehost the image if necessary.
Solved problems in spoilers.
1. Differentiate f(x) = (2e^x) - (e^-x)/7
I know that the answer is f'(x) = (2e^x) + (e^-x)/7, but I'm not clear on why. With terms separated by + or -, you just take the derivative of one + or - the derivative of the other, correct? And the first term falls under the "d/dx of e^x is e^x" rule, yes?
So with the second term, do I use the quotient rule?
quotient rule is (u'v-uv')/v^2
So I end up with something like [(e^-x)*7] / 49, right?
I know that the answer is f'(x) = (2e^x) + (e^-x)/7, but I'm not clear on why. With terms separated by + or -, you just take the derivative of one + or - the derivative of the other, correct? And the first term falls under the "d/dx of e^x is e^x" rule, yes?
So with the second term, do I use the quotient rule?
quotient rule is (u'v-uv')/v^2
So I end up with something like [(e^-x)*7] / 49, right?
2. f(x) = 2e^(3x+1) - 17
f'(x) = 6e^(3x+1)
Since f'(e^x) = f'(x) * e^x
f'(x) = 6e^(3x+1)
Since f'(e^x) = f'(x) * e^x
3. f(x) = ln (x^2+1)
f'(x) = 2x/(x^2 +1)
I know that d(lnx) = 1/x
Why is it 2x * 1/(x^2+1) and not 2x + 1/(x^2+1) ?
f'(x) = 2x/(x^2 +1)
I know that d(lnx) = 1/x
Why is it 2x * 1/(x^2+1) and not 2x + 1/(x^2+1) ?
4. A demand function is p = 400 - 2q, where q is the quantity of good sold for price $p. Which of the following is an expression for the total revenue, R, in terms of q?
I just too R = p*q and substituted the function for p to get R = (400-2q)*q Right?
5. The figure below* shows the graphs of marginal cost and marginal revenue. What production level could maximize the profit?
*The graph is a concave-up parabola labeled $/unit (MC) that intersects with a horizontal line labeled MR at x=1000 and x=3000
I hope that's clear, otherwise I can rehost the image if necessary.
TL DR on
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Maybe try thinking of it that way? (It's been a while since Calculus, but your denominator of 49 strikes me as wrong).
Quotient rule is (u'v - uv')/v^2
I think I've worked out problem #1, but this second one is stumping me.
As to the second one, you don't have a product to use with the Product Rule.
#2: f(x) = 2e^(3x+1) - 17
d/dx of (3x+1) = 3 and d/dx of 2e^x = 2e^x and d/dx of -17 is 0
So 3*2e^(3x+1) + 0 = 6e^(3x+1)
Still the d/dx of e^x = e^x rule.
Well, it works, but is completely unnecessary, so you're basically right; you use the quotient rule when it's something like differentiating (e^x)/x
d( a * e^-bx) = -b * a * e^-bx
so d ( (1/7)*e^(-1*x)) = (1/7) * (-1) * e^(-1*x))
For the second, it's more the rule of differentiating exponents -
d(e^u) = d(u) * e^u
(which is where your rule of d(e^x) = e^x comes from, since d(x) = 1
so
d(2e^(3x+1) - 17) =
d ( 2e^(3x+1)) - d(17) =
2 * d (e^(3x+1)) - 0 =
2 * d(3x+1) * e^(3x+1) =
2 * 3 * e^(3x+1)
*edit
hehe, took too long to type... oh well, no-one every complained for having too many answers
This seems sensible. I was just wary of taking the derivative of the exponent to equal 3 without some other calculus voodoo.
edit: thanks Gdiguy, that makes things much clearer. New problem to follow in OP
But using the quotient rule:
d/dx e^-x = -e^-x
d/dx 7 = 0
So
d/dx (e^-x)/7 =
((-e^-x * 7) - (e^-x * 0)) / 7^2 =
-e^-x * 7 / 7^2 =
-e^-x / 7
which is the correct answer.
Given two functions f and g, the derivative of the composite of these functions is:
d/dx [f(g(x))] = f'(g(x)) * g'(x)
So, in problem number 2, f(x) = 2e^x - 17 and g(x) = 3x + 1, so that
f(g(x)) = 2e^(3x + 1) - 17.
Now, taking f' and g' we see f'(x) = 2e^x and g'(x) is 3. So, evaluating f'(x) at x=3x+1:
d/dx[f(g(x))] = 2e^(3x+1) * 3 = 6e^(3x + 1)
f(x) = ln(x)
g(x) = x^2 + 1
s. t.
f(g(x)) = ln(x^2 + 1)
Derivatives:
f' = 1 / x
g' = 2x
=> f'(g(x)) * g'(x) = 1 / (x^2 +1) * 2X = 2x / (x^2 + 1)
which is your answer. There's no + 1 term since the derivative of the constant in g(x) is 0.
You need to use the chain rule here, one form mentioned earlier, another one that I use often is listed here:
Set u = x^2 + 1. Then f(x) = y = ln(u). So du/dx = 2x and dy/du = 1 / u.
So f'(x) = (dy/du)*(du/dx) = (1/u) * (2x) = 2x / (x^2 + 1).
And damn you for making me think about things like the Chain Rule again! =P
I'm posting the answers along with the problems. This is all review stuff that I want to be sure I understand.
Thanks again to everyone who has responded.
Alright, I'm not sure I completely understand the specifics behind the setup of this question, but through a little knowledge of standard Calc problems, this is what I gather they're looking for:
The demand function P(q) represents the price of each item as the total quantity sold changes (since it tells you they're sold at a price $P). It's a demand function, so as more and more units are sold, demand goes down, and people will pay less for them.
i.e.
P(0) = 400, or, before any others had been sold, the first unit had a price of $400.
P(100) = 200, or, after q=100 units were sold, the price of a unit was $200.
P(200) = 0, or, after q=400 units were sold, the price of a unit was $0.
So to find the total revenue, you want to add all these up (finding the area under the curve), and we can do this with our friend integration:
R(q) = ∫P(q) = ∫(400 - 2q) dq = 400q - q^2 + c
It sounds like you were given choices for this problem...is that one of them? I'm still not sure I understood the setup entirely correctly. :P
Where do you get that + c?
This particular question was multiple choice, but because I answered incorrectly I do not have access to the correct answer.
If i differentiate x^2+5 is get 2x+0
So if I integrate 2x i get x^2+C
C is a constant that could be anything because you don't know what that 0 was before you differentiated, maybe it was a 5, maybe it was something else.
This is what I was thinking as well when I read the question. Key to understanding the problem that f(q) is not the amount you make selling q goods, f(q) is the price people are willing to pay if you have already sold q goods (q is basically a simple way of indicating how scarce the good is). This is why algabraic substitution isn't going to work and instead you need to take the integral of the function.
Wendy's can't charge you 10 dollars for a hamburger because you're literally starving.
the total revenue at any given point would be 400q - 2q^2 for that problem. then to maximize that revenue you would take the derivative and set it equal to 0. in this cast that means your maximum total revenue is at 400 = 4q or q=100. the total revenue maximizing price is therefore $200.
This is what total revenue looks like.
then the question is wrong. it happens sometimes but when you give a specific term like 'total revenue' which has a very specific definition, it should have an accurate answer.
needless to say, integrating doesn't give you an answer that makes sense. if you take the answer given by integration as the formula for TR, you should be able to find the maximum total revenue by taking the derivative and setting it equal to zero. the derivative of 400q - q^2 is 400 -2q. set that equal to 0 and you get q = 200. at q = 200, p = 400 - 2(200) which = 0 this is wrong as it states that the revenue maximizing price is 0.
this is a very simple problem to test students that do exactly what was done here, assume that since its a calculus problem that the answer must be to integrate or derivitize. here it is not. If you give a calculus student a problem that states 'Y = mx + b, what is Y*x?' the answer is not to integrate y with respect to x.
also, I should address this as well.
R(q) = ∫P(q) = ∫(400 - 2q) dq = 400q - q^2 + c
this would actually be a definite integral that you integrate from 0 to q so the answer does not have a c in it. It would just be
As you said, in true economics the price cannot usually change quickly to fit demand, but the problem seems to imply the direct opposite to a student with no exposure to economics - they're giving you giving you the price as a function that decreases.
In a simple example where the price of each unit can change with each sale, integrating the demand function will give you the total revenue. You just add up the individual prices of each unit.
The reason you didn't get a meaningful result when you tried to maximize the result it is that there is no maximum or minimum in this case - the total revenue is a constant value. The demand function dictates what you charge for each item, and you integrate to find what you have after you've sold them all.
On the other hand, what you did in your fixed-price example was find a single fixed-price that maximizes revenue. That's more practical, but I think the question would have been more specific if that's what it was looking for. Unless this is from a Calc class that's geared toward Economics majors.
as for the 5th question: (don't read until you're stuck)
the only question is which point is the profit maximizing point? you have a graph that looks something like this but shifted to the right.
when you are at x = 1000, and you make 1 more unit what will happen? well, your marginal cost is less than your marginal revenue so your marginal profit is positive. when does it go negative? it goes negative at q=3001.
the profit maximizing production level is q=3000
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