SOLUTION: Graph and record the solution to the following linear programming problems.
1. A bakery with 5 bakers makes bread and cakes and sells them to local restaurants. Each loaf of bre
Question 560280: Graph and record the solution to the following linear programming problems.
1. A bakery with 5 bakers makes bread and cakes and sells them to local restaurants. Each loaf of bread requires 50 minutes of labor and ingredients costing $0.90 and can be sold for $1.20 in profit. Each cake requires 30 minutes of labor and ingredients costing $1.50 and can be sold for $4.00 profit. The bakers agree that no one will work more than 8 hours a day. They can spend no more than $190 per day on ingredients. How many loaves of bread and how many cakes should they make each day to maximize their profit? What is the maximum profit? Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! let x = number of loaves of bread
let y = number of cakes
you can convert all of your labor to hours as follows:
8 hours max in a day = 8 hours max in a day for each baker.
5 bakers times 8 hours a day equals a maximum of 40 hours total labor per day.
50 minutes of labor per loaf of bread = 50/60 = 5/6 hours of labor per loaf of bread.
30 minutes of labor per cake = 1/2 hours of labor per cake.
you can make a table of your constraint and profit equations as follows:
labor (hours) ingredients ($) profit ($)
loaf of bread 5/6 $.90 $1.20
cake 1/2 $1.50 $4.00
max l 40 $190.00 TBD
since the number of hours or the number of ingredients has to be greater than or equal to 0, you have 2 additional constraints as follows:
x >= 0
y >= 0
your objective function is:
profit = 1.2x + 4y
you graph your constraint equations and look for the corners of the region that satisfies the requirements of the equations.
you graph the equations by solving for y first and then graphing that equation.
the equation you graph is an equality rather than an ineguality.
you then apply rules of logic to determine the acceptable regions.
your first equation that you want to graph is:
(5/6)x + (1/2)y <= 40
change the equation to an equality for graphing purposes.
the equation becomes:
(5/6)x + (1/2)y = 40
subtract (5/6)x from both sides of the equation to get:
(1/2)y = 40 - (5/6)x
multiply both sides of the equation by 2 to get:
y = 2*(40 - (5/6)x) *****
your second equation that you want to graph is:
.9x + 1.5y <= 190
change the equation to an equality for graphing purposes.
the equation becomes:
.9x + 1.5y = 190
subtract .9x from both sides of this equation to get:
1.5y = 190 - .9x
divide both sides of this equation by 1.5 to get:
y = (190 - .9x) / 1.5 *****
the equation with ***** next to them are the equations you can graph.
the graph of both these equations is shown below:
since your region of acceptability is less than or equal to the values of the equations of each of these lines, then the controlling factor appears to be the number of hours that can be worked.
that is the region equal to or below the line of the equation of the labor and greater than x = 0 and y = 0.
that is the region bounded by x = 0 to 48 and y = 0 to 80.
your profit equation will be maximum or minimum at the intersection of the lines that bound the area of compatibility.
those intersections are:
(0,80)
(48,0)
when x = 0, y = 80
this means the following:
0 loaves of bread were made and 80 cakes were made.
80 cakes consumed 80 * 1/2 hours = 40 hours of labor.
80 cakes consumed 80 * 1.5 dollars = 120 dollars of ingredients.
both values of x and y met the constraint requirements.
the profit was equal to 0 * 1.20 + 80 * 4.00 = 320 dollars for the day
when x = 48, y = 0
this means the following:
48 loaves of bread were made and 0 cakes were made.
48 loaves of bread consumed 48 * (5/6) hours = 40 hours of labor.
48 loaves of bread consumed 48 * .9 dollars = 43.2 dollars of ingredients.
both values of x and y met the constraint requirements.
the profit was equal to 48 * 1.20 + 0 * 4.00 = 57.6 dollars for the day.
another intersection that meets the requirements is x = 0 and y = 0.
this particular intersection yielded 0 hours of labor and 0 dollars of ingredients with 0 profit and so was automatically discarded as not pertinent to the maximum profit equation since nothing could be lower than that.
in all cases, money was still available for ingredients but could not be used because of the limitation on the number of hours available.
