11(x-2)=6(x-3) so helppppppppppp
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Answer:
<em>The biggest profit that can be obtained by the shop owner= </em>IDR 2,750,000.
Step-by-step explanation:
We need to form a LPP( Linear Programming Problem) for the following problem and solve it to get the maximize the profit.
Let x denote the number of men's shoe and y denote the number of female shoes.
Maximize z= 10000x+5000y----------(1)
x≥100 ( since a shoe store owner wants to fill his shop with at least 100 pairs of men's shoes )
y≥150 (at least 150 pairs of women's shoes).
also x+y≤400 (The store can only accommodate 400 pairs of shoes).
x≤150 (male shoes should not exceed 150 pairs).
⇒ 100≤x≤150
Hence the optimal solution of an LPP always lie on the end points.
We have end points as:
(100,300), (150,250), (100,150),(150,150).
by putting these value in equation (1) we see which give the maximum solution.
for (100,300) i.e. x=100 and y=300: z=2,500,000
for (150,250) i.e. x=150 and y=250: z=2,750,000
for (100,150) i.e. x=100 and y=150: z=1,750,000
for (150,150) i.e. x=150 and y=150: z=2,250,000
Hence maximum profit is obtained at x=100 and y=300.
i.e. to maximize the profit:
<em>Number of men's shoes to be kept in store=100</em>
<em>and number of women's shoes to be kept in store=300</em>
<em>The biggest profit that can be obtained by the shop owner= </em>IDR 2,750,000.
