PLEASE HELP I'LL GIVE BRIANLIEST PLEASE THIS IS DUE PLEASE HELP ME PLEASE You’ve been given $500.00 for your fencing. So first we are going to make as big of a rectangular field as we can with the money that we have for the fencing to surround it. The cost of premium fencing (only the best for your farm, of course) is $11.41 per metre. So with that, and the following equations,
A=l*w
P=2l+2w
where A is the area, P is the perimeter, l is the length, and w is the width of a rectangle, we can find the maximum area that we can have for this field. You should encounter a quadratic function, so using a graphing calculator, or an online graphing website, will help you see what that looks like. You can then use methods you learn in the chapter to find the maximum of this function.


Now that you have a field with fencing around it, let’s grow some crops on it to make some money.
In the last section we used quadratics functions to maximize the area. Similarly, in this section we will be
using quadratic functions to maximize revenue.
Let’s say we have access to planting 2 different crops, Watermelons and Grapes. We can find the revenue of a crop by the equation:
R = (p0 + px)(n0 - nx)
Where R is the revenue, p0 is the starting price, n0 is the amount sold at the starting price, p is the price increase, n is the decrease in the amount sold per price increase, and x is the number of price increases. Crop A: Watermelons have a starting price of $12.00 per m^2 of area sold. At this price, you are able to sell 45 m^2 worth. For every $2.00 increase, you will sell 1 m^2 less. So the first task is to figure out how to put these values into the equation above, and turn it into a quadratic function. Then you will need to find the value of x in order to make the revenue as large as possible. Once you know x, the number of price increases, you can then find what you should price your Watermelons per m^2.