Question

A plane delivers two types of cargo between two destinations. Each crate of cargo I is 3 cubic feet in volume and 137 pounds in weight, and earns $30 in revenue. Each crate of cargo II is 3 cubic feet in volume and 274 pounds in weight, and earns $45 in revenue. The plane has available at most 270 cubic feet and 14,248 pounds for the crates. Finally, at least twice the number of crates of I as II must be shipped. Find the number of crates of each cargo to ship in order to maximize revenue. Find the maximum revenue.

crates of cargo I
crates of cargo II
maximum revenue		$

Answer 1

Let x crates of cargo I and y crates of cargo II be shipped. Each crate of cargo I and cargo II is 3 cubic feet in volume , and the plane has available at most 270 cubic feet so that 3x+3y ≤ 270 or, x+y ≤ 90…(1)

Also, each crate of cargo I weighs 137 pounds and each crate of and cargo II weighs 274 pounds and the plane has available at most 14,248 pounds for the crates so that 137x +274y ≤ 14248 or, x+2y ≤ 104…(2)

Further, since at least twice the number of crates of I as II must be shipped, hence x ≥ 2y…(3)

The revenue function is R(x) = 30x+45y. We have to maximize R(x) subject to the above 3 constraints.

The graph of the lines y = 90-x (in red)… (1), y = -x/2+52 ( in blue)…(2) and y = x/2( in green)…(3) is attached. The feasible region Δ ABC is in the 1^st quadrant ( as both x and y should be positive integers) on or below the 1^st ,2^nd and the 3^rd lines. Here A is the point of intersection of the 2^nd and the 3^rd lines, B is the point where the 3^rd line meets the X-Axis and C is the point where the 1^st line meets the X-Axis. Of these 3 points, we can easily rule out the points B and C as y cannot be 0. Now, at the point A, we have x = 52 and y = 26. Then R(x) = 30* 52+45*26 =1560+1170 = $ 2730.

Thus, for maximizing the revenue, we should have

crates of cargo I                : 52

crates of cargo II: 26

maximum revenue: $ 2730