This problem involves solving a system of inequalities (not to exceed, can be no more). To solve, we will graph equations and analyze the points of intersection. Let's write equations from the information given in the problem statement.....
Let:
a=book A
b=book B
6a+8b≤7200...Eq1, book cost not to exceed $7200
a≤560............Eq2, max number of books A printed
b≤720............Eq3, max number of books B printed
P=10a+b........Eq4, equation used to calculate profit
Note: All inequalities above are non-strict (≤), will be
graphed as solid lines.
Let's put Eq1 in slope-intercept form for graphing, solve for b.....
6a+8b≤7200
8b≤-6a+7200.........subtract 6a from both sides
b≤(-2/3)a+900.....Eq5, divide both sides by 8, simplify
From Eq5, m=-2/3........slope, coefficient of a, decreasing line
From Eq5, y-intercept=(0,900)
Let x-axis represent values of books "A"
Let y-axis represent values of books "B"
Now graph Eq5...shade region below the line
graph Eq2...shade region below the line
graph Eq3...shade region below the line
Consider portion of graph in Quadrant1 only, all positive values.
Intersection ALL REGIONS is the solution for this system of
inequalities. "Corner" coordinate pairs are critical points for
solution, are (0,0), (0,720), (240,720), (560,480), (560,0).
To evaluate the critical points, substitute into Eq4 for profit...
(0,0)..........P1=10(0)+0=0......................$0 profit
(0,270).......P2=10(0)+720=720...........$720 profit
(240,720)...P3=10(240)+720=3120....$3120 profit
(560,480)...P4=10(560)+480=6080....$6080 profit...maximum
(560,0).......P5=10(560)+0=560...........$560 profit
Therefore, using values from point (560,480), we conclude
that printing 560 copies of book A and 480 copies of book B
will give the publishing company maximum profit of $6080.
