- 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 + 4000we 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 constantnow 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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