Find the demand and supply equation

Question

When the price is 10, Demand 100, Supply 120.

When the price 7 Demand 130, Supply 90.

1)Find the linear price function.

2) Find the demand function.

3)Find the linear revenue function ( R(x) ) ?

4)If cost function ( (x) = 3x-5 max profit ?

These are my questions. I really need your help guys for my final exam.

Answer 1

It helps to try graphing it, using Price as the y-axis, and quantity demanded or supplied as the x-axis

The demand function is nearly always negatively sloped, as a higher price means less purchased.

You have two points for the demand function: (Q,P)=(100,10) and (Q,P)=(130,7) Think of those as (x,y) points on a 2 dimensional graph, with horizontal and vertical axes. As P or y increases from 7 to 10, Q or x decreases from 130 to 100. +3 change in y corresponds to a -30 change in x. The slope is "rise over run" or 3/(-30)=-1/10 The slope is also (y-10)/(x-100) for any (x,y).

set those two expressions for slope equal:

-1/10=(y-10)/(x-100) Cross multiply to get -x+100=10y-100

or x+10y=200 or Q+10P=200 as the linear demand curve or function

For supply: (7,90) and (10,120) are two points on the linear supply curve or function

That's (P,Q) or (x,y) points on a graphical straight line. It's slope is "rise over run" or (120-90)/(10-7)=30/3=10

supply curves are generally always upward sloping, with positive slope

The find the general slope using an arbitrary (x,y) point and one of the other points, such as (7,90)

slope = 10 = (y-90)/((x-7) cross multiply to get 10x-70=y-90 or 10x-y=-20

Revenue is price times quantity demanded or the demand function times price

Q+10P=200 or Q=200-10P

R=PQ=P(200-10P)=200P-10P²

Plot that graphically, and it's a parabola with a maximum revenue point at P=10. For either P>10 or P<10, revenue falls. Maximum revenue is 200(10)-10(10)²= 2000-1000=1000

3x-5 or 3Q-5 is the cost function. The revenue function is 200P-10P² or also PQ=(20-Q/10)Q= 20Q-Q²/10

Maximum revenue is when P=10 and Q=100

Profits are Revenue minus Costs = 20Q-Q²/10-3Q+5 = -Q²/10+17Q+5

Maximum profits occurs when -Q/5+17=0 or Q=85

Maximum profit is -(85)²/10+17(85)+5 = 722.5

Answer 2

1) To get linear price function: we have 2 points (100, 10) and (130, 7). Slope m = (7-10)/(130-100) = - 0.1.

Using poit-slope form p - 10 = - 01(x - 100), p = - 0.1x + 20

2) For supplay function we have 2 points (10, 120) & (7, 90). Slope m = ((7 - 10)/(90 - 120) = 0.1

Using point-slope form p - 10 = 0.1(x - 120), p = 0.1x - 2

3) Revenue function R(x) = px = (-0.1x + 20)x = -0.1x² + 20x

4) Profit function P(x) = R(x) - C(X), where C(X) = 3x - 5 is cost function.

Because P(x) = -0.1x² + 20x - 3x + 5, P(x) = -0.1x²+ 17x + 5.

Marginal profit P'(x) = -0.2x + 17. To get critical demand set P'(x) = -0.2x + 17 = 0, x = -17/(-0.2) = 85.

Maximum profit is P(85) = ·-0.1·85² +17·85 + 5 = 727.5