College Algebra Question 2?

Question

If anyone could help an English major out, I would really appreciate it :)

A manufacturer of widgets has fixed costs of $600 per month, and the variable cost is $60 per thousand widgets (so it costs $60 to produce a thousand widgets). Let N be the number, in thousands, of widgets produced in a month.

a. Find a formula for the manufacturer's total cost C as a function of N.

b. The highest price p, in dollars per thousand widgets, at which N can be sold is given by the formula p = 70 - 0.03N. Using this, find a formula for the total revenue R as a function of N.

c. Use your answers to parts a and b to find a formula for the profit P of this manufacturer as a function of N.

d. Use your formula from part c to determine the production level at which profit is maximized if the manufacturer can produce at most 300 thousand widgets in a month.

Thank you so m

Answer 1

OK, so the total cost is fixed + variable, so in this case:

C = 600 + 60N

The sales price per unit N is 70 -0.03N, so the total revenue realized from selling N units (that is, N thousands) is

R = N(70 - 0.03N)

Thus the profit, P (revenue - cost) is
P= N(70-0.03N) -(600 + 60N) or

P = 70N -0.03N^2 -600 -60N

or

P = -0.03N^2 + 10N -600

In order to find the profit maximum you have to differentiate the equation and look for the point at which R'(N) =0, which is either the maximum (in this case) or minimum of a quadratic. So,

R'(N) = 2*-0.03N + 10
or
R'(N)=-0.06N + 10

Now when R'(N) = 0,

0 = -0.06N +10
or

-10 = -0.06N

so N = -10/-0.06 = 166.67

and thus profit is maximized when N = 166 (not 167). So if the manufacturer can make at most 300,000 widgets a month then only 166 should be made.

Answer 2

Let N be number of widgets in thousands per month.

Then
Cost, C = 600 + 60N

Given also price, p = 70 - 0.03N
Total Revenue = Np = 70N - 0.03N^2

Profit, P = Revenue - Cost
P = 70N - 0.03N^2 - 600 - 60N
P = 10N - 0.03N^2 - 600

So for maximum profit, dP/dN = 0
dP/dN = 10 - 0.06N = 0
10 = 0.06N
So, N = 166.67 thousand units

So N = 166667 to nearest unit

Answer 3

a: C = 600 + 60N
b: R = (70 - 0.03N)N
c: P = (70 - 0.03N)N - (600 + 60N)
.. or. 10N - 0.03N^2 - 600
d: -0.03N^2 + 10N - 600 = 0
Using the quadratic formula, I get approx 255,000 and 78,000