Thread: pls help economics

Originally Posted by Mariana

Hi !

I really don't understant this problem. Can anyone solve this and help me to understand?

The relation between the price (p) per unit of a product and the number of units sold (x) is given by:

p=50-0.025x

a) Write down the expression for the total income function ()Ix
b) Write down the expression for the marginal income function.
c) For which value of x is the income at its maximum and what is the price in this situation?
d) Express the demand x(p) as a function of the price p.
e) Find an expression of the elasticity of the demand with respect to p.
f) For which value of p is the elasticity neutral?

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) Write down the expression for the marginal income function.

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).

I suppose you know how to find the derivative.
So I'(x) = 50 - 2(0.025)x
=> I'(x) = 50 - 0.05x

c) For which value of x is the income at its maximum and what is the price in this situation?

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) = 0

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

i'm not sure about (d), so i won't even try. if we find what part (d) should be, we should be able to figure out the other two questions, since the elasticity of demand is the measure of how demand changes as it responds to a change in price, that is delta(D)/delta(P), which we could find using another derivative. and the elasticity is nuetral where that function is zero

Hope i helped you at least some

Originally Posted by Mariana

Hi !

I really don't understant this problem. Can anyone solve this and help me to understand?

The relation between the price (p) per unit of a product and the number of units sold (x) is given by:

p=50-0.025x




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

The total income is the price times the sales, so:

In(x) = p(x) x = (50 - 0.025 x) x = 50 x - 0.025 x^2

b) Write down the expression for the marginal income function.

The marginal income function is the increase in income when production

is increased by one unit, so:

MI(x) = I(x+1) - I(x) = 50 (x+1) - 0.025 (x+1)^2 - 50 x + 0.025 x^2

........= (1999 - 2 x)/40

c) For which value of x is the income at its maximum and what is the price in this situation?

The income is at a maximum when the marginal income is zero, so in this case:

MI(x) = 0 = (1999 - 2 x)/40,

so: x = 1999/2

d) Express the demand x(p) as a function of the price p.
e) Find an expression of the elasticity of the demand with respect to p.

f) For which value of p is the elasticity neutral?