9514 1404 393
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.
Answer:
1 x 60, 2 x 30, 3 x 20, 4 x 15, 5 x 12, 6 x 10, 10 x 6, 12 x 5, 15 x 4, 20 x 3, 30 x 2, 60 x 1 = 60
Step-by-step explanation:
∫(t = 2 to 3) t^3 dt
= (1/4)t^4 {for t = 2 to 3}
= 65/4.
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∫(t = 2 to 3) t √(t - 2) dt
= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2
= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du
= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}
= 26/15.
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For the k-entry, use integration by parts with
u = t, dv = sin(πt) dt
du = 1 dt, v = (-1/π) cos(πt).
So, ∫(t = 2 to 3) t sin(πt) dt
= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt
= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]
= 5/π + 0
= 5/π.
Therefore,
∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.
678.58 that is the surface area to this problem
Answer:
8
Step-by-step explanation:
The perimeter is the sum of the 3 sides of the triangle.
let the other 2 sides be x and x - 3 , then
12 + x + x - 3 = 31
2x + 9 = 31 ( subtract 9 from both sides )
2x = 22 ( divide both sides by 2 )
x = 11 and x - 3 = 11 - 3 = 8
The 3 sides are 12, 11, 8
with shortest side 8
