A store sells two types of toys, A and B. The store owner pays $8 and $14 for each one unit of toy A and B respectively. One unit of toys A yields a profit of $2 while a unit of toy B yields a profit of $3. The store owner estimates that no more than 2000 toys will be sold every month and he does not plan to invest more than $20,000 in inventory of these toys. How many units of each type of toys should be stocked in order to maximize his monthly total profit (12 marks)
Answers
Answer:
Let x be the total number of toys A and y the number of toys B; x and y cannot be negative, hence
x≥0 and y≥0
The store owner estimates that no more than 2000 toys will be old every month.
x+y≤2000
One unit of toys A yields a profit of $2 while a unit of toys B yields a profit of $3, hence the total profit P is given by
P=2x+3y
The store owner pays $8 and $14 for each one unit of toy A and B respectively and he does not plant to invest more than $20,000 in inventory of these toys
8x+14y≤20,00
What do we have to solve?
Find x and y so that P=2x+3y is maximum under the conditions
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⎪
⎪
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⎪
⎪
⎪
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⎪
⎪
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x≥0
x≥0
x+y≤2000
8x+14y≤20,000
The solution set of the system of inequalities given above and the vertices of the region obtained are shown below:
Vertices of the solution set
A at (0,0)
B at (0.1429)
C at (1333,667)
D at (2000,0)
Calculate the total profit P at each vertex
P(A)=2(0)+3(0)=0
P(B)=2(0)+3(1429)=4287
P(C)=2(1333)+3(667)=4667
P(D)=2(2000)+3(0)=4000
The maximum profit is at vertex C with x=1333 and y=667
Hence the store owner has to have 1333 toys of type A and 667 toys of type B in order to maximize his profit
solution