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.
I did my math and I think 41 minuets but are u adding or subracting
9514 1404 393
Answer:
65.6%
Step-by-step explanation:
The question is essentially asking the ratio of the volumes.
Va = πr²h = π(15²)(20)
Vb = πr²h = π(19²)(19)
The ratio of volumes is ...
Va/Vb = π(15²)(20)/(π(19³)) = (15²)(20)/19³ ≈ 0.65607
Container B will be about 65.6% full when container A is empty.
Solution
f(x) = -20
+14x + 12 and g(x) = 5x - 6.
(f/g)(x) = 
(f/g)(x) = 
<u>Step 1: </u>Now we have to factorize the numerator.
f(x) = -20x^2 + 14x + 12
Factor out -2, we get
= -2 (10x^2 - 7x - 6)
Now we can factorize 10x^2 - 7x - 6
f(x) = -2(2x + 1) (5x - 6)
<u>Step 2: </u>Plug in the factors
(f/g)(x) = 
<u>Step 3:</u> Cancel out the common factor (5x - 6) from the numerator and the denominator, we get
(f/g)(x) = -2(2x +1) = -4x -2
Since -4x -2 is linear expression, the domain is all the real numbers.
Therefore, the answer is –4x – 2; all real numbers
Thank you :)
(8,1);((-2,-6)
2•8-3•1>=12
16-3>=12
13>=12
2•(-2)-3(-6)>=12
-4+18>=12
14>=12
