Pre-calculus problem?

Question

A real estate developer plans to build a new apartment complex on an empty lot. He knows that if he builds atwenty-unit complex he will be able to charge $600/month for rent. He also knows that if his apartment complex only has sixteen (slightly larger) units he will be able to charge $700/month for rent instead.

1. Assume that the relationship between the number of units in the complex and rent per month is linear. Write a formula for rent charged as a function of the number of units.
2. The total monthly rental revenue would be the product of the number of units times the rent per unit, and the developer would like to maximize this. How many units should he build to do this? What could he charge for rent? What will his monthly revenue be?

(Detailed explanation needed please)

Answer 1

Call the number of units 'x.'
We want to find a function of 'x' (call it 'f(x)') which gives the amount of rent the developer can charge.
We know: f(20) = 600 and f(16) = 700.
The relationship is linear, so the formula will follow the familiar "y=mx+b" format. The slope is rise/run; i.e., (700-600)/(16-20) = -25.
So f(x) = -25x + b.
Since f(20) = 600 = -25*20 + b, b = 1100.
So the formula is f(x) = -25x + 1100.

The total monthly revenue is the number of units multiplied by the revenue per unit; i.e., x*f(x).
So this formula is -25x² + 1100x, a parabola opening downward. The vertex of the parabola is the optimization point.

Answer 2

the following is the thanks to sparkling up utilizing row-decrease x + y + z = 7 3x - 2y + z = 3 x + 6y + 3z = 25 a million a million a million | 7 3 -2 a million | 3 a million 6 3 | 25 R2 = R2 - 3 * R3 a million a million a million | 7 0 -20 -8 | -seventy 2 a million 6 3 | 25 R3 = R3 -R1 a million a million a million | 7 0 -20 -8 | -seventy 2 0.5 2 | 18 R3 = R3 + a million/4 * R2 a million a million a million | 7 0 -20 -8 | -seventy 2 0 0 0 | 0 R1 = R1 + a million/20 * R2 a million 0 .6 | 3.4 0 -20 -8 | -seventy 2 0 0 0 | 0 R2 = -R2/20 a million 0 .6 | 3.4 0 a million .4 | 3.6 0 0 0 | 0 So the finest constraints are that: x + .6z = 3.4 y + .4z = 3.6 case in point, ideas comprise: x = a million a million + .6z = 3.4 .6z = 2.4 z = 4 y + .4z = 3.6 y + a million.6 = 3.6 y = 2 x = a million, y = 2, z = 4 x = 2 2 + .6z = 3.4 .6z = a million.4 z = 2.33.. y + .4(2.33..) = 3.6 y = 2.sixty six x = 2, y = 2.sixty six, z = 2.33

Answer 3

Yes, exactly what they said.

To maximize the total revenue for the complex, use:

let R= total revenue
let u= number of units

R = u(-25u+1100)

The vertex of the parabola is the optimization point.