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Sagot :
Answer:
[tex]\$1320[/tex]
Step-by-step explanation:
Let [tex]x[/tex] be the number of units of A
[tex]y[/tex] be the number of units of B
For assembling we have
[tex]20x+25y\leq 3000[/tex]
For packaging we have
[tex]10x+5y\leq 1200[/tex]
Let profits earned be [tex]Z[/tex] so
[tex]Z=10x+8y[/tex]
We have the maximize the function.
Plotting the equations we can see that the intersection points are [tex](0,120),(100,40),(120,0)[/tex]
In the question it is mentioned we have to sell both the products so [tex]x[/tex] or [tex]y[/tex] cannot be [tex]0[/tex].
So, the point of maximiztion for the function would be [tex](100,40)[/tex]
The maximum profit would be
[tex]Z=10\times 100+8\times 40\\\Rightarrow Z=1320[/tex]
![View image boffeemadrid](https://us-static.z-dn.net/files/dae/ca7a91aeb26a9e57b9f7dea29568cd8f.png)
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