Solve the equation d/3=11
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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
(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
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.
From here, differentiate the problem.
Now, set it equal to zero and solve for x.
This a critical point of the function. Let's plug back into the original equation to see what it gives us.
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
Plug back into P(x) to see the profit
Now, do the same for 163 plots
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.