This is quite an interesting problem. I am not sure how high you are in math, but I am going to use calculus I techniques to solve it. First, we need to model an equation. Let P be the total profit and x be every time you increase the cost by $10. If you think about it hard enough you come up with the equation
[tex]P(x)=(200-5x)(250+10x)[/tex]
(200-5x) is the amount of plots you will be able to sell, and (250+10x) is the amount you charge for. So, at x =0
[tex]P(0)=(200-5(0))(250+10(0))=(200)(250)=$50,000[/tex]
This is the initial condition where if we sell 200 plots at $250/plot.
So, this equation makes sense.
Now, let's maximize using the first derivative of the function.
Let's get it into an easily differentiable form.
[tex]P(x)=(200-5x)(250+10x)=-50x^2+2000x-1250x+50000\\=-50x^2+750x+50000[/tex]
From here, differentiate the problem.
[tex]P'(x)=-100x+750[/tex]
Now, set it equal to zero and solve for x.
[tex]P'(x)=-100x+750=0\\x=7.5[/tex]
This a critical point of the function. Let's plug back into the original equation to see what it gives us.
[tex]P(7.5)=(200-5(7.5))(250+10(7.5))=(162.5)(325)=52,812.50[/tex]
You cant sell half a plot, so we need to see what happens if we sell 162 plots and 163 plots, and then compare which one gives us more money.
In order to sell 162 plots
[tex]200-5x=162\\x=7.6[/tex]Plug back into P(x) to see the profit
[tex]P(7.6)=(200-5(7.6))(250+10(7.6))=(162)(326)=52,812[/tex]
Now, do the same for 163 plots
[tex]200-5x=163\\x=7.4\\P(7.4)=(200-5(7.4))(250+10(7.4))=(163)(324)=52,812[/tex]
As we can see, they are the same. So, you can charge either $324 or $326 in rent. But, if your teacher is not looking for a logical answer and you can somehow sell half a plot, you can charge $325 in rent for the maximum profit.
