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.
The answer is: No
Becuase the equation itself is not correct. The equation is c = 2p.
Good Luck!
Standard form : -5a+75
leading co is: -5
Answer:
y = x - 4
General Formulas and Concepts:
<u>Algebra I</u>
Slope-Intercept Form: y = mx + b
- m - slope
- b - y-intercept
Step-by-step explanation:
<u>Step 1: Define</u>
Slope <em>m</em> = 1
y-intercept <em>b</em> = -4
<u>Step 2: Write function</u>
y = x - 4
<h3>
Answer: ds/dt = 11</h3>
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Work Shown:
Before we can use derivatives, we need to find the value of s when (x,y) = (15,20)
s^2 = x^2+y^2
s^2 = 15^2+20^2
s^2 = 225+400
s^2 = 625
s = sqrt(625)
s = 25
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Now we can apply the derivative to both sides to get the following. Don't forget to use the chain rule.
s^2 = x^2 + y^2
d/dt[s^2] = d/dt[x^2 + y^2]
d/dt[s^2] = d/dt[x^2] + d/dt[y^2]
2s*ds/dt = 2x*dx/dt + 2y*dy/dt
2(25)*ds/dt = 2(15)*5 + 2(20)*(10)
50*ds/dt = 150 + 400
50*ds/dt = 550
ds/dt = 550/50
ds/dt = 11
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Side note: The information t = 40 is never used. It's just extra info.
