Answer to Question #174091 in Operations Research for Mimi

Question

when Tracy McCoy wakes up Saturday morning, she remembers

that she promised the PTA she would make some cakes and/or homemade bread for its bake sale

that afternoon. However, she does not have time to go to the store to get ingredients, and she has

only a short time to bake things in her oven. Because cakes and breads require different baking

temperatures, she cannot bake them simultaneously, and she has only 3 hours available to bake.

A cake requires 3 cups of flour, and a loaf of bread requires 8 cups; Tracy has 20 cups of flour.

A cake requires 45 minutes to bake, and a loaf of bread requires 30 minutes. The PTA will sell a

cake for $10 and a loaf of bread for $6. Tracy wants to decide how many cakes and loaves of

bread she should make.

a. Formulate a linear programming model for this problem.

b. Solve this model by using graphical analysis.

Answer 1

Solution:

Let the number of cakes be $x$ and number of loaves of bread be $y$ .

Objective function: To maximise cost, $Z=10x+6y$

Subject to the constraints:

$3x+8y\le20 \\ 45x+30y\le3(60)\Rightarrow3x+2y\le12 \\x,y\ge0$

Consider them equations and plotting on the graph, we get following graph.

Also, put (0,0) in these inequations.

$0\le20\Rightarrow True \\0\le12\Rightarrow True$

Thus, their shadow or shaded area is towards (0,0).

Now, OABC is the feasible region.

Solving these equations to get point B.

$O(0,0),A(4,0),B(\frac {28}9,\frac 43),C(0,2.5)$

At $O(0,0): Z=0+0=0$

At $A(4,0):Z=10(4)+0=40$

At $B(\frac {28}9,\frac 43):Z=10(\frac {28}9)+6(\frac 43)\approx31.1+12=43.1$

At $C(0,2.5):Z=0+6(2.5)=15$

Clearly, maximum is 43.1 at B.

Hence, she should make $\frac {28}9$ cakes and $\frac 43$ loaves of bread.