Answer:
[tex]x\geq0, y\geq0, x+y\geq20,[/tex] and [tex]8x+13y\leq300[/tex].
Step-by-step explanation:
Let x and y be the numbers of the paperback book and the hardcover book respectively.
as x and y are the counting numbers, so it will be always positive. So,
[tex]x\geq0\;\dots(i)[/tex] and
[tex]y\geq0\;\cdots(ii)[/tex]
Given that the weight of 1 paperback book=0.8 pounds and
the weight of 1 paperback book=1.3 pounds.
There are at least 20 books in each box, so
[tex]x+y\geq20\;\cdots(iii)[/tex]
As one box cannot weigh more than 30 pounds, so
[tex]0.8x+1.3y\leq30[/tex]
[tex]\Rightarrow 8x+13y\leq300\;\cdots(iv)[/tex]
Now, from the equations (i),(ii),(iii), and (iv), all the sets of constraints as
[tex]x\geq0, y\geq0, x+y\geq20,[/tex] and [tex]8x+13y\leq300[/tex].
Moreover, the sets of all the integral values of (x,y) in the shaded region are the feasible numbers of the paperback book and hardcover book.
