You asked this question already here http://www.mathhelpforum.com/math-he...272-s-o-s.html

Please don't double post. Anyway, let's see which of these i can do.

a) Write down the expression for the total income function ()Ix

a) The total income is the money we make, that is, it is the price times the amount of units we sell. the number of units sold is x, the price, p, is given by p = 50 - 0.025x. So the income function I(x) is given by:

I(x) = x*p = x(50 - 0.025x) = 50x - 0.025x^2

b) if my limited knowledge of economics serves me well, I think the marginal income is the income gained from selling one more unit of good. So that would be how the income changes when we increase the value of x by some number. Now i may be wrong, but that, to me, sounds like a derivative. we want to see how I(x) changes as x changes, so we want the derivative of I with respect to x, that is, I'(x) = the marginal income function. (I feel more confident about that guess now, i remember someone mentionaing that to me before, the marginal curve is always the derivative of the cost or income curve--let's hope they were right).b) Write down the expression for the marginal income function.

I suppose you know how to find the derivative.

So I'(x) = 50 - 2(0.025)x

=> I'(x) = 50 - 0.05x

we need to find the value of x where the income function is at it's maximum. this will be where the derivative, that is, the marginal function I'(x) = 0c) For which value of x is the income at its maximum and what is the price in this situation?

set I'(x) = 0

=> 50 - 0.05x = 0

=> -0.05x = -50

=> x = 50/0.05

=> x = 1000

so when 1000 units are sold, the income is maximized. but what is the price at this point?

p = 50-0.025x

when x = 1000

=> p = 50 - 0.025(1000) = 50 - 25 = 25

so the price at this point is $25