The dance teacher bought a roll of ribbon that is 201.5 yards. She needs 15.5 yards of ribbon for each dancer. How many 15.5 yar
1 answer:
You might be interested in
Answer:
35.4 years
Step-by-step explanation:
The annual consumption (in billions of units) is described by the exponential function ...
f(t) = 45.5·1.026^t
The accumulated consumption is described by the integral ...
We want to find t such that the value of this integral is 2625, the estimated oil reserves.
2625 = 45.5/ln(1.026)·(1.026^t -1)
2625·ln(1.026)/45.5 +1 = 1.026^t ≈ 1.480832 +1 = 1.026^t
Taking natural logs, we have ...
ln(2.480832) = t·ln(1.026)
t ≈ ln(2.480832)/ln(1.026) ≈ 35.398
After about 35.4 years, the oil reserves will run out.
A) Profit is the difference between revenue an cost. The profit per widget is
m(x) = p(x) - c(x)
m(x) = 60x -3x^2 -(1800 - 183x)
m(x) = -3x^2 +243x -1800
Then the profit function for the company will be the excess of this per-widget profit multiplied by the number of widgets over the fixed costs.
P(x) = x×m(x) -50,000
P(x) = -3x^3 +243x^2 -1800x -50000
b) The marginal profit function is the derivative of the profit function.
P'(x) = -9x^2 +486x -1800
c) P'(40) = -9(40 -4)(40 -50) = 3240
Yes, more widgets should be built. The positive marginal profit indicates that building another widget will increase profit.
d) P'(50) = -9(50 -4)(50 -50) = 0
No, more widgets should not be built. The zero marginal profit indicates there is no profit to be made by building more widgets.
_____
On the face of it, this problem seems fairly straightforward, and the above "step-by-step" seems to give fairly reasonable answers. However, if you look at the function p(x), you find the "best price per widget" is negatve for more than 20 widgets. Similarly, the "cost per widget" is negative for more than 9.8 widgets. Thus, the only reason there is any profit at all for any number of widgets is that the negative costs are more negative than the negative revenue. This does not begin to model any real application of these ideas. It is yet another instance of failed math curriculum material.
m(x) = p(x) - c(x)
m(x) = 60x -3x^2 -(1800 - 183x)
m(x) = -3x^2 +243x -1800
Then the profit function for the company will be the excess of this per-widget profit multiplied by the number of widgets over the fixed costs.
P(x) = x×m(x) -50,000
P(x) = -3x^3 +243x^2 -1800x -50000
b) The marginal profit function is the derivative of the profit function.
P'(x) = -9x^2 +486x -1800
c) P'(40) = -9(40 -4)(40 -50) = 3240
Yes, more widgets should be built. The positive marginal profit indicates that building another widget will increase profit.
d) P'(50) = -9(50 -4)(50 -50) = 0
No, more widgets should not be built. The zero marginal profit indicates there is no profit to be made by building more widgets.
_____
On the face of it, this problem seems fairly straightforward, and the above "step-by-step" seems to give fairly reasonable answers. However, if you look at the function p(x), you find the "best price per widget" is negatve for more than 20 widgets. Similarly, the "cost per widget" is negative for more than 9.8 widgets. Thus, the only reason there is any profit at all for any number of widgets is that the negative costs are more negative than the negative revenue. This does not begin to model any real application of these ideas. It is yet another instance of failed math curriculum material.