Question #174091

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.


Expert's answer

Solution:

Let the number of cakes be xx and number of loaves of bread be yy .

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

Subject to the constraints:

3x+8y≤2045x+30y≤3(60)⇒3x+2y≤12x,y≥03x+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≤20⇒True0≤12⇒True0\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(289,43),C(0,2.5)O(0,0),A(4,0),B(\frac {28}9,\frac 43),C(0,2.5)

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

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

At B(289,43):Z=10(289)+6(43)≈31.1+12=43.1B(\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)=15C(0,2.5):Z=0+6(2.5)=15

Clearly, maximum is 43.1 at B.

Hence, she should make 289\frac {28}9 cakes and 43\frac 43 loaves of bread.


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