Question

A local bakery sells cakes and pies. The cakes take 2 hours of preparation time and 1 hour to decorate. The pies take 3 hours of preparation time and 0.5 hours to decorate. The profit is $25 per cake and $12 per pie. The bakery can spend at most 108 hours on preparation and 20 hours on decorating their items per day.

a.Write the system of linear inequalities to represent this situation.
b.Sketch the feasible region.
c.Write the optimization equation.
d.Determine the maximum profit. Show all work.

Answer 1

a) Let's assume C represents the number of cakes sold and P represents the number of pies sold.

The inequality for preparation time is:
2C + 3P ≤ 108 (since cakes take 2 hours and pies take 3 hours to prepare)

The inequality for decoration time is:
1C + 0.5P ≤ 20 (since cakes take 1 hour and pies take 0.5 hours to decorate)

The non-negativity constraints are:
C ≥ 0 (the number of cakes cannot be negative)
P ≥ 0 (the number of pies cannot be negative)

b) To sketch the feasible region, we need to plot the graph of each inequality and shade the area where the inequalities are satisfied.

For the preparation time:
2C + 3P ≤ 108
We can rewrite this as:
P ≤ (108 - 2C) / 3

For the decoration time:
1C + 0.5P ≤ 20
We can rewrite this as:
P ≤ 40 - 2C

Now, let's plot the graphs:

First, draw the line P = (108 - 2C) / 3.
Then draw the line P = 40 - 2C.

The feasible region is the region where both inequalities are satisfied, which is below both lines.

c) The optimization equation is the equation that represents the objective we want to maximize. In this case, the objective is to maximize profit. The profit equation can be written as:

Profit = 25C + 12P

d) To determine the maximum profit, we need to find the corner point within the feasible region that maximizes the profit equation.

Let's find the intersection of the two lines:

(108 - 2C) / 3 = 40 - 2C
108 - 2C = 3(40 - 2C)
108 - 2C = 120 - 6C
4C = 12
C = 3

Substituting C = 3 into P = (108 - 2C) / 3:
P = (108 - 2(3)) / 3
P = (108 - 6) / 3
P = 102 / 3
P = 34

The intersection point is (C, P) = (3, 34).

Now, substitute these values into the profit equation:
Profit = 25C + 12P
Profit = 25(3) + 12(34)
Profit = 75 + 408
Profit = 483

So, the maximum profit is $483.