Agrocer sells milk chocolate at $2.50 per pound, dark chocolate at $4.30 per pound, and dark chocolate with almonds at $5.50 per pound. he wants to make a mixture of 50 pounds of mixed chocolates to sell at $4.54 per pound. the mixture is to contain as many pounds of dark chocolate with almonds as milk chocolate and dark chocolate combined. how many pounds of each type must he use in this mixture?

10 pounds of milk chocolate, 15 pounds of dark chocolate, and 10 + 15 = 25 pounds of almond chocolate

Step-by-step explanation:

Let x be the number of pounds of milk chocolate that he needs to use in the mixture. And y is the number of pounds of of dark chocolate. Since the mixture must use almonds chocolate as much as the other 2 combined in term of weight, and the total weight is 50 pounds. That means

milk_choco_weight + dark_choco_weight + almond_choco_weight = 50

x + y + (x + y) = 50

2(x+y) = 50

x+y = 25 or x = 25 - y

Since we know the price of the mixture, we can use the following equation

2.5milk_choco_weight + 4.3dark_choco_weight + 5.5almond_choco_weight = 4.54total_weight

2.5x + 4.3y + 5.5(x+y) = 50*4.54

2.5x + 4.3y + 5.5*25 = 227

2.5x + 4.3y = 227 - 137.5 = 89.5

We can substitute x = 25 - y

2.5(25 - y) + 4.3y = 89.5

62.5 - 2.5y + 4.3y = 89.5

1.8y = 27

y = 15

so x = 25 - y = 25 - 15 = 10

So we need 10 pounds of milk chocolate, 15 pounds of dark chocolate, and 10 + 15 = 25 pounds of almond chocolate

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

i dont know sorry

step-by-step explanation: