MATH SOLVE

7 months ago

Q:
# 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 m2 of area sold. At this price, you are able to sell 45 m2 worth. For every $2.00 increase, you will sell 1 m2 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 m2.
Equation:
Number of price increases: x =
R=
Number of price increases: x =
Price of watermelons per m^2 =

Accepted Solution

A:

To turn the given information into the quadratic function for the revenue of Watermelons, we can substitute the given values into the formula:
R = (p0 + px)(n0 - nx)
where p0 = $12.00, n0 = 45, p = $2.00, and n = 1.
Substituting these values, we get:
R = (12 + 2x)(45 - x)
Expanding the brackets, we get:
R = 540 + 18x - 2x^2
So the quadratic function for the revenue of Watermelons is R = -2x^2 + 18x + 540.
To find the value of x that maximizes revenue, we can use the formula for the x-value of the vertex of a quadratic function:
x = -b / (2a)
In this case, a = -2 and b = 18, so:
x = -18 / (2*-2)
x = 4.5
So the optimal number of price increases for Watermelons is 4.5. Since we can't have a fractional number of price increases, we'll round up to 5.
To find the optimal price of Watermelons per m2, we need to add 5*$2.00 = $10.00 to the starting price of $12.00, giving us a final price of $22.00 per m2.
Therefore, the optimal price for Watermelons per m2 is $22.00, and the optimal number of price increases is 5.