Rainforest data:
1 answer:
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Answer:
150 quartz of creamy vanilla and 50 quartz of Continental Mocha.
Step-by-step explanation:
Let x represents the number of quartz of Creamy vanilla and y represents the number of Continental Mocha,
∵ There is a profit of $ 3 on each quart of Creamy Vanilla and $2 on each quart of Continental Mocha.
So, the total profit,
Z = 3x + 2y
Which is the objective function for this problem that has to maximise,
Each quart of Creamy Vanilla go 2 eggs and 3 cups of cream,
So, in creamy vanilla,
Number of eggs = 2x
Cups of cream = 3x
Also, Each quart of Continental Mocha go 1 egg and 3 cups of cream,
So, in creamy vanilla,
Number of eggs = y
Cups of cream = 3y
Thus, total number of eggs = 2x + y
And, total number of cups of cream = 3x + 3y
According to the question,
2x + y ≤ 350
3x + 3y ≤ 600
Which are subject of constraint,
Graphing :
Related equation of 2x + y ≤ 350 is 2x + y = 350,
Having x-intercept = (175,0) y-intercept = (0, 350)
Also, 2(0) + 0 ≤ 350 ( True )
∴ Shaded region of 2x + y ≤ 350 would contain the origin,
Related equation of 3x + 3y ≤ 600 is 3x + 3y = 600,
Having x-intercept = (200,0) y-intercept = (0, 200)
Also, 3(0) + 3(0) ≤ 600 ( True )
∴ Shaded region of 3x + 3y ≤ 600 would contain the origin,
'≤' shows solid line,
By graphing them we obtain a feasible region,
In which boundary points are (0, 200), (150, 50) and (175,0)
At (0, 200),
Z = 3(0) + 2(200) = $ 400,
At ( 150, 50)
Z = 3(150) + 2(50) = 450 + 100 = $ 550
At ( 175, 0)
Z = 3(175) + 2(0) = $ 525
Hence, 150 quartz of creamy vanilla and 50 quartz of Continental Mocha should make to earn the largest profit.
