a candy company has 130 pounds of cashews and 170 pounds of peanuts which they combine into different mixes the deluxe has half cashews and half peanuts and sells for $7 a pound the economy mix has one third cashews and two thirds peanuts and sells for 5.40 a pound how many pounds of each mix should be prepared for maximum revenue

If you have x lbs of deluxe and y lbs of economy, then you want to

maximize p=7x+5.4y subject to
x/2 + y/2 <= 130
2x/3 + y/3 <= 170

now graph the lines and find the vertex with maximum p.

I still don't get is I knopw max of parabola is -b/2a but this isn't quadratic

To determine the quantities of each mix that should be prepared for maximum revenue, we need to set up a system of equations based on the information given.

Let's assume x represents the number of pounds of the deluxe mix, and y represents the number of pounds of the economy mix.

Given that the deluxe mix has half cashews and half peanuts, we can determine the amount of cashews and peanuts in this mix:
Cashews in deluxe mix: 0.5x
Peanuts in deluxe mix: 0.5x

Similarly, the economy mix has one third cashews and two thirds peanuts, so we can determine the amount of cashews and peanuts in this mix:
Cashews in economy mix: (1/3)y
Peanuts in economy mix: (2/3)y

To find the total revenue, we multiply the price per pound by the weight of each mix:
Revenue from deluxe mix: 7 * x
Revenue from economy mix: 5.40 * y

We also have the constraint that the total weight of the mixes should not exceed the available quantities of cashews and peanuts:
Total cashew constraint: 0.5x + (1/3)y ≤ 130
Total peanut constraint: 0.5x + (2/3)y ≤ 170

To maximize revenue, we need to formulate an objective function. In this case, it is the total revenue:
Maximize Revenue: Revenue from deluxe mix + Revenue from economy mix = 7x + 5.40y

We can now set up the linear programming problem:
Objective function: Maximize 7x + 5.40y
Subject to constraints:
0.5x + (1/3)y ≤ 130
0.5x + (2/3)y ≤ 170

Solving this linear programming problem will give us the optimal quantities of each mix to maximize revenue.

To calculate the maximum revenue, we need to determine how many pounds of each mix to prepare.

Let's assume the number of pounds for the deluxe mix is x, and the number of pounds for the economy mix is y.

For the deluxe mix, it has half cashews and half peanuts, so we can allocate x/2 pounds of cashews and x/2 pounds of peanuts.

For the economy mix, it has one third cashews and two thirds peanuts. Therefore, we can allocate y/3 pounds of cashews and 2y/3 pounds of peanuts.

To determine the maximum revenue, we need to consider the prices. The deluxe mix sells for $7 per pound, while the economy mix sells for $5.40 per pound.

The total revenue can be calculated by multiplying the pounds of each mix by their respective prices and adding them together.

Revenue = (x/2) * $7 + (y/3) * $5.40

To find the maximum revenue, we need to optimize this revenue function by finding the maximum value. One way to do this is by using calculus.

To find the maximum revenue, we need to take the derivative of the revenue function with respect to both x and y, and set them equal to zero. Then we can solve for x and y.

d(revenue)/dx = (7/2)x - 0 = 3.5x
d(revenue)/dy = 5.4/3y - 0 = 1.8y

Setting these derivatives equal to zero:

3.5x = 0
1.8y = 0

Solving these equations, we find x = 0 and y = 0. It means that we have no revenue when x = 0 and y = 0 pounds.

This doesn't seem right, so let's reassess the problem. It appears that there is a constraint missing. Without any constraints, the maximum revenue would be obtained by preparing an infinite amount of each mix.

Could you please check if there is any additional information or constraints given in the problem?