Please answer

Question

At a unit price of $55 , the quantity demanded of a certain commodity is 1000 units. At a unit price of $85, the demand drops to 600 units. A) given that it is linear , find the demand equation. B) graph the demand equation. C)above which price will there be no demand? D) what is the maximum quantity demanded? E) what is the quantity demanded when the unit price is $65? F) what is the slope of the demand equation? What dose the slope mean in terms of the quantity demanded and dollars?

Answer 1

Hi Marc,

 

We'll take each part of our question in turn below.

 

 

Part A:

 

Let's begin by writing the slope-intercept form of a linear equation:

    y = mx + b (m is slope; b is y-intercept)

 

To calculate the slope (m), we will use the slope formula:

    m = (y₂ - y₁) / (x₂ - x₁)

 

We'll plug in the coordinates of our two given points:

    (x₁,y₁) = (55,1000)

    (x₂,y₂) = (85,600)

    m = (1000 - 600) / (55 - 85)

    m = -(400/30)

    m = -(40/3)

 

To calculate b (y-intercept), we can plug into the slope-intercept form of the linear equation the values for our slope (m) and the coordinates of either one of our given points.  We'll use the coordinates of our first given point:

    y = mx + b (slope-intercept form)

    1000 = -(40/3)(55) + b (our plug-ins)

    1000 = -(2200/3) + b

    b = 1000 + 2200/3

    b = 3000/3 + 2200/3

    b = 5200/3

 

Now we can write our demand equation in slope-intercept form with our values for slope (m) and y-intercept (b):

    y = -(40/3)x + 5200/3

 

 

Part B:

 

Below is the link to the graph of our demand equation:

 

https://dl.dropbox.com/s/eyls2pt5vuong92/Graph_of_Demand_Equation.png?raw=1

 

The graph shows our two given points (55,1000) and (85,600) as well the other points in answer to the remaining parts of our question.

 

 

Part C:

 

The price above, which there will be NO demand is the x-intercept of our graph (where y = 0).  We can calculate it as follows by setting y to zero in our demand equation and solving for x:

    y = -(40/3)x + 5200/3

    -(40/3)x + 5200/3 = 0

    -40x + 5200 = 0

    -40x = -5200

    (-1)(-40x) = (-1)(-5200)

    40x = 5200

       x = 5200/40

       x = 520/4

       x = $130 (NO demand above this price)

 

This value is confirmed by the x-intercept point shown on our graph (130,0).

 

 

Part D:

 

The maximum quantity demanded is where the price is zero, which is the y-intercept of our graph or the value of b in the slope-intercept form of our demand equation:

    y = mx + b (slope-intercept form)

    y = -(40/3)x + 5200/3 (demand equation)

    b = 5200/3

    b = 1733 and 1/3

    b = 1733 units

          (maximum demand rounded down to

              nearest whole unit)

 

This value is confirmed by the y-intercept point shown on our graph (0,1733.3).

 

 

Part E:

 

To find the quantity demanded when the unit price is $65, we plug in 65 for x and solve for y using our demand equation:

    y = -(40/3)x + 5200/3 (demand equation)

    y = -(40/3)(65) + 5200/3

    y = 5200/3 - 2600/3

    y = 2600/3

    y = 866.666

    y = 867 units

          (demand rounded up to

              nearest whole unit)

 

This point is also shown on our graph (65,866.66)

 

 

Part F:

 

The slope of the demand equation is our calculated value for m in answer to Part A:  

      m = -(400/30)  [actual ratio]

       m = -(40/3)       [ratio reduced to lowest terms] 

 

In terms of demand quantity and unit price, it means that for every increase of $30 in the unit price, there is a corresponding decrease in demand of 400 units.

 

Thanks for submitting this problem and glad to help.

 

God bless, Jordan.