Pythagoreans Pizza, Inc. is a makes amazing deep dish pizza. The company makes a profit of $1.00 for each cheese pizza and $1.50 for each sausage pizza produced. Currently, Pythagoreans Pizza has 150 pounds of dough mix and 50 pounds of topping mix. Each cheese pizza uses 1 pound of dough mix and 4 ounces (16 ounces= 1 pound) of topping mix. Each sausage pizza uses 1 pound of dough mix and 8 ounces of topping mix. Based on the past demand per week, Pythagoreans Pizza can sell at least 50 cheese pizzas and at least 25 sausage pizzas. Determine the number of cheese and sausage pizzas the company should make to maximize profit.
 
 

 

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The above graph shows the constraints. The maximum profit is given by the point (100,50), indicating 50 sausage pizzas and 100 cheese pizzas, profit $175. The enclosed region R satisfies all constraints.

EXPLANATION

Let C=number of cheese pizzas and S=the number of sausage pizzas. The red line represents the amount of dough available. Each pizza uses a pound of dough so C+S≤150, and it’s the area below the red line that tells us that we can use no more than 150lbs. Similarly the blue line represents topping: 0.25C+0.5S≤50.

The area to the right of the green line tells us we must sell at least 50 cheese pizzas and the area above the pink line tells us we must sell at least 25 sausage pizzas.

All these constraints put together give us region R which contains all possible combinations of the two types of pizza. The red and blue lines intersect at the solution of two equations: C+S=150 and 0.25C+0.5S=50. The second equation can be written C+2S=200 by multiplying all the terms by 4. By subtracting the first equation from this we get 2S-S=200-150=50, so S=50 and C=100. This happens to be where the profit equation Profit=C+1.5S is maximum, so we have 100+75=175 when we put C=100 and S=50. No other points in R give us a higher profit. Therefore, 50 sausage pizzas and 100 cheese pizzas is the answer.

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