Hi alexinia!
Here are the answers to your questions.
Question 1
For this and the following questions, recall that the marginal revenue
and marginal cost are the derivatives of the revenue and cost
functions respectively.
In this question,
R(x) = 20 x - .005x^2
The derivative is
R'(x) = 20 - 0.01x
Evaluating at x = 1000, we get:
R'(1000) = 20 - 0.01*1000 = 10
Thus the marginal revenue when 1000 tables are sold is 10
Question 2
We have
g(x) = sqrt(x)*f(x)
Using the multiplication rule for derivatives, we get that:
g'(x) = 1/(2*sqrt(x))*f(x) + sqrt(x)*f'(x)
Evaluating at x = 4, we get
g'(4) = 1/(2*sqrt(4))*f(4) + sqrt(4)*f'(4) = 3/4 + 2*5 = 10.75
Question 3
The slope of this tangent line should be precisely the derivative of
f(x) evaluated at the point where the tangent line passes. Since the
line touches x=-3, and f'(-3)=5, then the slope of the line should be
5. Therefore, the line is of the form:
y = 5x + b
We also know that this line passes through the point (-3,2) (because
f(-3)=2), so we can use this information to find b:
2 = 5*(-3) + b
2 = -15 + b
b = 17
Therefore, the equation of the tangent line is 5x + 17.
Question 4
The logic here is the same as in question 2. We know how to use the
division rule for derivatives:
g(x) = f(x)/x
g'(x) = (f'(x)*x - f(x))/x^2
g'(4) = (f'(4)*4 - f(4))/4^2 = -23/16
Question 5
This is a two-part function. We have:
/ 7x - 2 x ? 1
F(x) = |
\ kx^2 x > 1
Clearly, F(x) is continuos for x<1 (because 7x-2 is continuous) and is
continuous for x>1 (because kx^2 is continuous, no matter the value of
k). So the only "conflictive" point is x=1. In order for F to be
continuous the value of F should be the same at x=1 either if we
approach x=1 from the left (thus using 7x-2) or if we approach x=1
from the right (thus using kx^2).
So, when x = 1, then 7x - 2 = 5. Also, when x = 1, kx^2 = k.
Therefore, in order for F to be continuous over all of its domain, we
must set k = 5. Thus F(x) will tend to 5 if we approach x=1 from the
left or the right.
Question 6
There appears to be something wrong with the equation provided here.
Could you please confirm if it's the following through a clarification
request? I understand that what you wrote is:
120
C(X) = 3x^2 + ------ + 1
2x
but this doesn't seem right (why wouldn't you just put 60 instead of
120/2?). Please clarify this using the clarification request feature
and I'll answer it as soon as possible.
In case you want to check that the procedure you followed for this
question was right, this is what should be done:
First, from the cost function we're given, we must find the average
cost function. This is done dividing C(x) by x. Once we have this
function C(x)/x, we must take its derivative with respect to x, and
that's the marginal average cost.
Question 7
The cost function here is
C(x) = 100 + 15x - x^2
The instantaneous rate of change of cost with respect to the number of
cases produced is just the marginal cost. Therefore, we find the
derivative and then evaluate it at x = 1.
C'(x) = 15 - 2x
C'(1) = 13
Therefore, the instantaneous rate of change of cost with respect to
the number of cases when just one case is produced is 13.
Question 8
Although you wrote here "Re: Question 6", it appears that you meant
"Re: Question 7", since the question is about "cases", that appear
also in the latter. Please request a clarification otherwise.
In this case, we must just evaluate the marginal cost function at x = 5.
C'(5) = 15 - 2*5 = 5
So the marginal cost when 5 cases are produced is 5.
Question 9
We have the following revenue and cost functions:
R(x) = 6x ? (x^2 /100)
C(x) = 2x
The marginal cost function is the derivative of the cost function. In this case,
C'(x) = 2
is the marginal cost function
Question 10
Again, here you mention "Re: Question 8" when this question apparently
refers to question 9. I'll assume so, please let me know through a
clarification request if otherwise. This happens in question 11 too.
The marginal revenue is the derivative of the revenue function:
R'(x) = 6 - x/50
Question 11
We must first find the marginal profit function. We have that:
Marginal Profit = Marginal Revenue - Marginal Cost
Marginal Profit = 6 - x/50 - 2
Marginal Profit = 4 - x/50
Now, we have to find the value of x that makes the marginal profit equal to 0.
0 = 4 - x/50
x/50 = 4
x = 200
So, when 200 units are produced, marginal profit equates to zero.
Finally, we must find the profit when x = 200. We have that:
Profit = Revenue - Cost
Profit = 6x ? (x^2 /100) - 2x
Profit = 4x - (x^2 /100)
So, when x = 200,
Profit = 4*200 - (200^2/100) = 400
Therefore, when marginal profit is 0, profit is 400, and 200 units are
being produced.
Google search terms
marginal revenue
://www.google.com.ar/search?hl=es&q=marginal+revenue&meta=
marginal cost
://www.google.com.ar/search?hl=es&q=marginal+cost&meta=
I hope this helps! Please remember to request clarification for
question 6 in order to confirm the apparently wrong equation, so I can
answer it as soon as possible.
Best wishes!
elmarto |
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