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Answer:
- Constraints: x + y ≤ 250; 250x +400y ≤ 70000; x ≥ 0; y ≥ 0
- Objective formula: p = 45x +50y
- 200 YuuMi and 50 ZBox should be stocked
- maximum profit is $11,500
Step-by-step explanation:
Let x and y represent the numbers of YuuMi and ZBox consoles, respectively. The inventory cost must be at most 70,000, so that constraint is ...
250x +400y ≤ 70000
The number sold will be at most 250 units, so that constraint is ...
x + y ≤ 250
Additionally, we require x ≥ 0 and y ≥ 0.
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A profit of 295-250 = 45 is made on each YuuMi, and a profit of 450-400 = 50 is made on each ZBox. So, if we want to maximize profit, our objective function is ...
profit = 45x +50y
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A graph is shown in the attachment. The vertex of the feasible region that maximizes profit is (x, y) = (200, 50).
200 YuuMi and 50 ZBox consoles should be stocked to maximize profit. The maximum monthly profit is $11,500.
Make an equation for this find the variables P and t and 200. Them plug the into an equation like y=mx+b.
Answer:
db / dt = kb
this becomes b(t) = Ce^(kt)
C = 100, the initial population
P(1) = 420 = 100 e^(1k)
4.2 = e^k
ln 4.2 = k
a) thus, b(t) = 100 e^(t ln 4.2)
b) b(3) = 100 e^(3 ln 4.2)
c) growth constant will still be ln 4.2 (constant percentage of populatioin)
d) 10000 = 100 e^(t ln 4.2)
100 = e^(t ln 4.2)
ln 100 = t ln 4.2
t = ln 100 / ln 4.2
Step-by-step explanation:
Answer:
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Step-by-step explanation:
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