- JuliaZ 1 year
you know that at $400, there is 100 unit rented out,
so Revenue = 100 * 400.
you also know as rent increase by 5, unit decrease by 1.
we can rewrite the equation as,
Revenue = (100 – n)(500 + 5*n)
n for each $5 increase.
Expanding the equation gives
R = -5x^2 + 100n + 4000
we can use this equation to find the derivative of each term
so derivative of -5x ^ 2 is -5(2)x
derivative of 100n is 100
derivative of 4000 is 0 because it is a constant
now combining the derivatives, the derivative of revenue
is -10n + 100. when at the maximum revenue, the tangent must be 0. so
-10n + 100 = 0
n = 10.
n is the number of $5 increases.
so manager should charge 400 + 5(10) = 450 to maximize revenue.
The manager knows 100-unit apartment complex will be occupied if the rent is $400 per month
A market survey suggests that, on the average, one additional unit will remain vacant for each $5 increase in rent. What rent should the manager charge to maximize revenue?
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