The Pit BBQ food truck sells two combo packs of ribs and brisket. Combo A contains 10 oz of brisket and 16 oz of ribs and sells for $21.25. Combo B contains 20 oz of brisket and 8 oz of ribs and sells for $30.00. They plan to use no more than 400 oz of brisket and 384 oz of ribs and they want to sell at most 28 combo packs. How many of each combo should be made to maximize revenue ?

They should make of Combo A and of Combo B which would earn them $ in revenue.

Based on recent changes in meat prices, The Pit has decided to change the prices to $29.75 for Combo A and $27.20 for Combo B. In this case, they should instead make of Combo A and of Combo B to earn $ in revenue.

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Raphealnwobiid Raphealnwobiid · Answer 1

They should make 19 of Combo A and 10 of Combo B which would earn them $703.75 in revenue.

Let x represent the number of combo A and y represent the number of combo B. Hence:

10x + 20y ≤ 400 (1)

Also:

16x + 8y ≤ 384 (2)

The solution to the inequality is (19, 10)

Revenue = 21.25(19) + 30(10) = $703.75

They should make 19 of Combo A and 10 of Combo B which would earn them $703.75 in revenue.

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