Total number of chocolate boxes that can be produced: x+y (<,<=,>,>=) ___ restrictions based on demand of each: y(<,<=,>,>=) ___x
maximum amount of white chocolate production: y(<,<=,>,>=) _
minimum amount of milk chocolate production: x(<,<=,>,>=) _
minimum amount of white chocolate production: y(<,<=,>,>=) _
vertices of feasible region : (0,0)(400,___)(_,___)(___,0)
optimization equation: profit = _x+_y
your maximum profit is $_ .you should produce _ boxes of milk chocolate and _ boxes of white chocolate .

Step-by-step explanation:

The idea here is to create lines according to the constraints we were given, graph the lines (which are actually inequalities), and then shade in the region that satisfies the inequality. Let's start at the beginning of the problem and we'll get our lines (inequalities) written.

The total number of boxes that can be produced according to the constraints is 800, so the inequality for that is

x + y ≤ 800 and solving for y:

y ≤ 800 - x

Another constraint on the white chocolate is that it has to be less than or equal to 200 boxes, so:

y ≤ 200

The max number of white chocolate boxes is half the number of milk chocolate, so:

y ≤ (1/2)x

The min number of milk chocolate boxes produced is:

x ≥ 0 and

The min number of white chocolate boxes produced is:

y ≥ 0 (This means that it is a possibility of making 0 milk chocolate boxes and all white chocolate boxes OR there is a possibility of making 0 white chocolate boxes and all milk chocolate boxes)

The production equation (which is used later) is:

2.25x + 2.50y (you make a profit of $2.25 on every milk chocolate box you sell and profit of $2.50 on every white chocolate box you sell).

The bold equations are the ones that need to be graphed (see graph below). Where those 3 lines intersect are the vertices of feasible region:

(0, 0), (400, 200), (600, 200), (800, 0).

Then take each x and y value from a coordinate and plug it into the profit equation (we don't need to use (0, 0)) starting with x = 400 and y = 200:

2.25(400) + 2.5(200) = $1400

Now using x = 600 and y = 200:

2.25(600) + 2.5(200) = $1850

Now using x = 800 and y = 0:

2.25(800) + 2.5(0) = $1800

So our max profit as seen by the evaluations is $1850, and that occurs when we sell 600 boxes of milk chocolate and 200 boxes of white chocolate.

395.84

step-by-step explanation:

2πrh+2πr²

2 (pie)(radius)(height)+2(pie)(radius)^2