Read the following scenario and develop a method for answering the question posed. Be sure to define all variables used, justify your thinking mathematically, and fully answer the questions posed in complete sentences. Orbital Toys sells two types of sets of magnetic spheres, silver, and brass. The store owner, Lucy Ball, pays $8 and $16 for each one set of silver magnetic spheres and brass magnetic spheres respectively. One set of silver magnetic spheres yields a profit of $5 while a set of brass magnetic spheres yields a profit of $7. Ms. Ball estimates that no more than 2000 sets of magnetic spheres will be sold every month and she does not plan to invest more than $20,000 in the inventory of these sets. How many sets of each type of magnetic spheres should be stocked in order to maximize her total monthly profit? What is her maximum monthly profit?

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EndiaW248894 EndiaW248894 · Answer 1 · 2022-10-25T14:52:09+02:00

8 dlls for silver

16 dlls for brass

5 dlls profit silver

7 dlls profit spheres

Then the price is

8+5= 13 dlls for silver

16+7=23 dlls for brass

Let S and B be the amount of magnetic silver and brass sphers that are sold, respectively.

Then, Ms. Ball estimation is that

[tex]S+B\leq2000[/tex]

Also, she doesn't want to invest more than 20000, so

[tex]\begin{gathered} 8S+16B\leq20000 \\ S+2B\leq2500 \end{gathered}[/tex]

The objective function is

[tex]V=5S+7B[/tex]

Subjected to:

[tex]\begin{gathered} S+B\leq2000 \\ S+2B\leq2500 \\ S\ge0,\text{ B}\ge0 \end{gathered}[/tex]

GRAPH

The interection is at

[tex]\begin{gathered} S=2000-B \\ S=2500-2B \\ 2000-B=2500-2B \\ B=500 \\ S=2000-500 \\ S=1500 \end{gathered}[/tex]

So, the extremes must be at (0,1250), (1500,500), (2000,0) , (0,0).

So, if we replace the points

[tex]\begin{gathered} V(0,1250)=5(0)+7(1250)=8750 \\ V(1500,500)\text{ = 5(1500)+7(500)=}11000 \\ V(2000,0)=5(2000)+7(0)=10000 \end{gathered}[/tex]

So, the amount she will need to stock to maximize her profit is 1500 of silver and 500 of brass, and the maximum profit is going to be 11000 dlls.