Answer
30 orange crates, 30 grapefruit crates
Work Step by Step
Define x as the number of orange crates and y as the number of grapefruit crates. Since the trucker profits 2.50 dollars per orange crate and 4 dollars per grapefruit crate, we want to maximize 2.50x + 4y.
Since each orange crate is $4ft^{3}$ and each grapefruit crate is $6ft^{3}$, x orange crates and y grapefruit crates occupy 4x + 6y cubic feet. As the truck has a maximum of 300 cubic feet, 4x + 6y $\leq$ 300.
Each orange crate is 80 pounds, and each grapefruit crate is 100 pounds, so x orange crates and y grapefruit crates weigh 80x + 100y pounds. As the truck holds a maximum of 5600 pounds, 80x + 100y $\leq$ 5600.
The trucker cannot hold haul grapefruit crates than orange crates, so y $\leq$ x. We also must have y $\geq$ 0 and x $\geq$ 0, since we can't have a negative amount of crates. Finally, we then graph these constraints:
Since maximums occur at corners, we take the coordinates of the corners of the feasible area, which are (0,0), (30,30), (45,20), and (70,0). Then, we plug their x and y values into 2.50x + 4y and find the maximum:
(0,0): 2.50(0) + 4(0) = 0
(30,30): 2.50(30) + 4(30) = 195
(45,20): 2.50(45) + 4(20) = 192.5
(70,0): 2.50(70) + 4(0) = 175
Since the maximum occurs at (30,30), the trucker should bring 30 orange crates and 30 grapefruit crates.
Since the maximum occurs at
