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1. A retiree has 60 garden plots on which he could plant cabbage and pechay. His friend told him that he could make a profit of Php 100 per slot of cabbage and Php 90 per slot of pechay. His household could not possibly take care of more than20 plots of cabbage and 40 plots of pechay. How many plots of each vegetable should he plant for him to have a maximum profit? What is the maximum profit? 2. A chemist makes a component of orchid fertilizer that requires two secret ingredients labeled as A and B. She sells ingredient A at Php 200 per kilogram and ingredient B at Php 350 per kilogram. Per production, a kilogram of ingredient A requires 4 units of special pollen while 3 units for each kilogram of ingredient B. Suppose the excess of ingredient B over ingredient A per production should be at most 4 units, and there are only 24 units of special pollen available per production, how many kilograms of ingredient A and ingredient B should be made per production to maximize the sales?

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Jianne Marie Tapang


Quiz 6.2 Linear Programming
1. A retiree has 60 garden plots on which he could plant cabbage and pechay. His friend told
him that he could make a profit of Php 100 per slot of cabbage and Php 90 per slot of
pechay. His household could not possibly take care of more than20 plots of cabbage and 40
plots of pechay. How many plots of each vegetable should he plant for him to have a
maximum profit? What is the maximum profit?

Let x – number of plots of Cabbage
Let y – number of plots of Pechay
Objective Function: Plot the Constraints:
To maximize: P = 100x + 90y Find: x ≤ 20 Find: x + y ≤ 60
Constraints: x = 20 x + (40) = 60 (20) + y = 60
x ≤ 20 (20, 0) x = 60-40 y = 60-20
y ≤ 40 x = 20 y = 40
x + y ≤ 60 Find: y ≤ 40 (20, 40)
x ≥ 0, y ≥ 0 y = 40
(0, 40)




Values of the
Extreme Points Objective
Function:
(20, 0) 100(20) + 90(0) = The retiree should plant 20 plots of cabbage
Php 2,000 and 40 plots of pechay for him to have a
(0, 40) 100(0) + 90(40) = maximum profit. The maximum profit he will
Php 3,600 have is a total of Php 5,600.
(20, 40) 100(20) + 90(40) =
Php 5,600

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