A confectioner sells two types of nut mixtures. The standard-mixture package contains 100 g of cashews and 200 g of peanuts and sells of $1.95. The deluxe-mixture package contains 150 g of cashews and 50 g of peanuts and sells for $2.25. The confectioner has 15 kg of cashews and 20 kg of peanuts available. On the basis of past sales, the confectioner needs to have at least as many standard as deluxe packages available. How many bags of each mixture should she package to maximize her revenue? For this problem we will let x represent the number of standard-mixture packages and let y represent the number of deluxe-mixture packages. The values of x and y that will maximize her revenue depend on the amount of cashews and peanuts available for use. These are known as constraints as they constrain the possible values of the variables. 1. Thereare15kgofcashewsavailableandsincethestandard-mixtureuses100gandthedeluxe-mixtureuses150gwemust make sure that 100x+150y 15,000 2x+3y 300. 2. There are 20 kg of peanuts available and since the standard-mixture uses 200 g and the deluxe-mixture uses 50 g we must make sure that 200x+50y 20,000 4x+y 400. 3. The problem also states that she needs to have at least as many standard as deluxe packages available so we need xy yx. 4. Lastly, since we cannot produce a negative number of packages it must be that x0 and y0. The goal of this problem is to find the values of x and y that maximize her revenue. Since the standard-mixture sells for $1.95 and the deluxe-mixture sells for $2.25, her revenue will be given by the equation R = 1.95x + 2.25y. If we plot the constraints on the same set of axes, the ordered-pair, (x,y), that maximizes her revenue will be a vertex of the 2 I do not need to see the work. Can you just give the answer to Part 1 # 1,2,3 and Part 2 # 1, 2 thank you I have tried to figure this out