Question 180879
A real estate office handles an apartment complex with 50 units. When the rent
 per unit is $580 per month, all 50 units are occupied. However, when the rent
 is $625 per month, the average number of occupied units drops to 47.
 Assume that the relationship between the monthly rent p and the demand x is linear.
:
a. Write the equation of the line giving the demand x in terms of the rent p.
:
Using the slope formula: m = {{{(y2-y1)/(x2-x1)}}} find the slope
Assign the given values as follows
x1 = 580; y1 = 50
x2 = 625; y2 = 47
m = {{{(47-50)/(625-580)}}} = {{{(-3)/(45)}}} = {{{-1/15}}}
Write the equation using the point/slope formula y - y1 = m(x - x1)
y - 50 = {{{-1/15}}}(x - 580)
y - 50 = {{{-1/15}}}(x - 580)
y = {{{-1/15}}}x + 38.67 + 50
y = {{{-1/15}}}x + 88.67
The way they have written it, demand (d) dependent on price (p) it would be
d = {{{-1/15}}}p + 88.67
:
:
b. Use a graphing utility to graph the demand equation and use the trace feature to estimate the number of units occupied when the rent is $655. Verify your answer algebraically.
:
Substitute 655 for p in the above equation
d = {{{(-1/15)}}}(655) + 88.67
d = -43.67 + 88.67
d = 45 units rented at $655
:
 On a TI83, it would look something like this:
(Xmin=-100, Xmax=800; Ymin=-20, Ymax=100)
{{{ graph( 300, 200, -100, 800, -20, 100, (-1/15)x+88.67) }}}
:
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c. Use the demand equation to predict the number of units occupied when the rent is lowered to $595. Verify your answer graphically.
: 
Substitute 595 for p in the above equation in the same way
d = {{{(-1/15)}}}(595) + 88.67
d = -39.67 + 88.67
d = 49 units rented at $595, verify on the same graph
:
:
they should not have assigned the demand as x. x is usually the independent
variable and here, the demand is the dependent variable. Causes confusion.
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Hopefully, this has relieved some of your frustration, and you can enjoy the rest of this blessed Sunday!  Carl