Question

a firm can produce at most 100 units each week. if its total cost function is known to be C(x)= 500 + 1500x with a total revenue of R (x) = 1600x + x^2, how many units, x, should the firm produce to maximize its profit? Find the max. profit

Answer 1

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Answer 2

profit = revenue - cost =
\[1600x+x^2-(500+1500x)
\]

Answer 3

=
\[x^2-1500x+1100\] if my algebra is right yes?

Answer 4

oh it is R (x)= 1600x - x^2

Answer 5

well then it makes more sense

Answer 6

so we get
\[-x^2-1500x+1100\] yes?

Answer 7

wait my algebra is wrong

Answer 8

\[1600x-x^2-(500+1500x)\]

Answer 9

this time i get
\[-x^2+100x-500\] and now i think it is right. sorry

Answer 10

maximum is at
\[x=-\frac{b}{2a}=-\frac{100}{-2}=50\]

Answer 11

so firm should produce 50 units and its profit will be whatever you get when you replace x by 50

Answer 12

yeah sorry my bad
50 into original equation?

Answer 13

yes. i get 2000

Answer 14

\[-50^2+100\times 50 -500=2000\]

Answer 15

but you should check it because i used a calculator

Answer 16

profit function = 100x + x^2 - 500. differentiate this equation with respect to x and set equal to zero. Then solve for x. you want ur marginal profit from selling an extra unit to be zero.

Answer 17

profit function is 100x -x^2 - 500. sorry

Answer 18

differentiating is not necessary. vertex of a quadratic is
\[-\frac{b}{2a}\] whether you
a) take the derivative, set = 0 and solve,
b) complete the square, or
c) remember how to get a vertex

Answer 19

certainly do not need calc for this problem

Answer 20

yeah i got 2000
so max profit?

Answer 21

from an economic perspective it makes more sense to want marginal profit zero. but either way obviously works

Answer 22

so 2000 is max profit?

Answer 23

yes. produce 50, get a profit of 2000