An agricultural company has 80 tons of Type I fertilizer and 120 tons of Type II fertilizer. The company mixes both types of fertilizer into 3 products. Product X requires 3 parts of Type I and 1 part of Type II. Product Y requires 1 part of Type I and 4 parts of Type II. Product Z requires half a part of Type I and 2 parts of Type II. Product X is sold at $2000 per ton, Product Y at $1500 per ton, and Product Z is sold at $2500 per ton. Use the following decision variables: x: number of tons to mix of Product X
y: number of tons to mix of Product Y
z: number of tons to mix of Product Z

Requried:
Formulate the company's mathematical optimization problem.

$527.7.if the following was the question  $527.7 is the answer.students who are trying to raise funds have an agreement with a local pizza chain. the chain has agreed to sell them pizzas at a discount, which the students can then resell to families in the local community for a profit. it is expected that of the 500 families in the community, at most 70% will buy pizza. based on a survey of their personal preferences, the students believe that they should order no more than 120 cheese pizzas, no more than 150 pepperoni pizzas, and no more than 100 vegetarian pizzas. they also want to make sure that at least 20% of the total pizzas are cheese and at least 50% of the pizzas are pepperoni. they make a profit of $1.45, $1.75, and $1.98, respectively, for each cheese, pepperoni, and vegetarian pizza they resell. how many pizzas of each type should they buy?

Money believe but i may be wrong

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