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K Brown owns a large department building consisting of 200 units. From past experience, it is known that all 200 units can be rented out at $800 per month per unit and only 180 units can be rented if the monthly rent is $1000 per month. Assuming that the demand for apartments is a LINEAR FUNCTION of the monthly rent charged per unit, determine the total revenue AS A FUNCTION of monthly rent per unit. What monthly rent per unit will maximize the total revenue?

Answer 1

So let x = rent and y = demand
You have two points: (800, 200) and (1000, 180)

Given the linear assumption:
The equation to use is (y - y1) = (y2 - y1)/(x2 - x1) * (x - x1)

So: (y-200) = (180-200)/(1000-800) * (x-800)
(y-200) = -0.1(x-800)
y - 200 = -0.1x + 80
y = -0.1x + 280

Now, let T = total revenue.
The total revenue is the demand multiplied by the rent, ie:

T = xy
T = x(-0.1x + 280)
T = -0.1x^2 + 280x
(relationship between total revenue T and rent x)

Now, to find the value of x for which T is a maximum:
dT/dx must be equal to 0

dT/dx = -0.2x + 280 = 0
0.2x = 280
x = 280/0.2 = 1400

So the monthly rent per unit which will maximize the total revenue is $1400.

Answer 2

So, if
R = rent
N = number of units rented (i.e. demand)

then
N = a * R + b
R = (N - b)/a
with a and b being constants.

You have
200 = a * 800 + b
180 = a * 1000 + b
which you can solve for a, b (a will be negative)

For the total revenue Rev:
Rev = N * R = (a * R +b) * R = a * R² + b * R

The second part is:
Rev -> Max
for 0 <= N <= 200 -> -(b/a) >= R >= (200 - b)/a

Hmm, that should be helpful enough, now do the math.
*sigh* Guess I'm the only one who believes in hints only?

Answer 3

If you charge $800 you rent out all 200, if you charge $200 more you rent out 20 less, so the gradient of the LINEAR FUNCTION is -0.1units per dollar

If you had unlimited units and could charge no rent you'd rent out 280 units, so this is the intercept of the LINEAR FUNCTION.

so... the LINEAR FUNCTION can be expressed as:
Number rented (R)= - 0.1P + 280
(where R= number rented and P=Price per month)

so... Income (I)= RP= (-0.1)P^2 + 280P

The maximum of this function occurs when P=1400

(as dI/dP = -0.2P + 280 therefore dI/dP= 0 when 0.2P = 280)

The maximum revenue is 1400 * 140 units = $196000

(to check, try P=1200, 1200*160 = 192000 and P=1600, 1600*120= 192000, both of which are less than the maximum of $196000)

Answer 4

(u - 200) / (r - 800) = (180 - 200) / (1000 - 800)
u - 200 = - 20(r - 800) / 200
u = - 0.1r + 280
R = ru
R = r(- 0.1r + 280)
R = - 0.1r^2 + 280r
R = - 0.1(r^2 - 2,800r + 1,960,000) + 196,000
R = - 0.1(r - 1,400)^2 + 196,000
A rent of $1,400 per unit will bring in a revenue of $196,000 from 140 units