Answer:
$5,760
Explanation:
This is a problem of linear programming on two variables which can be solved graphically.
Let x ≥ 0 and y ≥ 0 the number of units of compact and standard machines respectively.
Restrictions
“The store can sell up to 90 copiers a month”
x+y ≤ 90
(See picture 1 attached)
“A compact copier requires 6 cu. ft. of storage space, and a standard copier requires 18 cu. ft. and maximum of 1080 cubic feet of storage space is available.”
6x + 18y ≤ 1080
(See picture 2 attached)
“The compact and standard copy machines take, respectively, 1 and 1.5 sales hours of labor. A maximum of 99 hours of labor is available.”
1x + 1.5y ≤ 99
(See picture 3 attached)
Vertex V1 = (18, 54) is the intersection of the lines
6x +18y = 1080
x + 1.5y = 99
which is the solution of the linear system of equations.
Vertex V2 = (72, 18) is the intersection of the lines
x + y = 90
x + 1.5y = 99
We want to find the maximum profit, i.e., the maximum of
P = 60x +80y
on the region obtained by intersecting all the above regions.
Since P is a linear function, the maximum is attained on one of the vertexes
(0, 60), (18, 54), (72, 18), (90, 0)
Let's evaluate P at each of this points to find the maximum out.
P(0, 60) = 80*60 = $480
P(18, 54) = 60*18 + 80*54 = $5,400
P(72, 18) = 60*72 + 80*18 = $5,760
P(90, 0) = 60*90 = $5,400
And we can see that the maximum is obtained when 72 units of compact machines and 18 units of standard machines are sold, and the maximum profit is $5,760.