let
a = journal
b = globes
Since the stand can only contain 100 newspapers,
![a+b\leq100](https://img.qammunity.org/2023/formulas/mathematics/college/aviiljxkpdi92e5aby3twbo240e8u5cqzg.png)
The constraint will be
![\begin{gathered} a+b\leq100 \\ a\ge20 \\ b\ge25 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/eh75di2pilc0nwz0n39sxwijb7r69lzylg.png)
To maximize profits
![\text{p}=0.05a+0.10b](https://img.qammunity.org/2023/formulas/mathematics/college/c041pgddjv4ay2lhnra6c3dvlfttt7jok4.png)
let us represent it with a graph. The corner point (vertices) will be
(20, 25)
(20, 80)
(75, 25)
Therefore,
The maximum profit will be
![\begin{gathered} \text{p}=0.05a+0.10b \\ \text{p}=0.05(20)+0.10(80)=1+8=\text{ \$9} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/iw0axdzzm3n0uvbthmpdna7ybn9o0x9cac.png)
He should sell 20 journals and 80 globes to get the maximum profit($9)