Answer:
$11,728
Step-by-step explanation:
Twice a year, for 4 years is 8 times
516×8 = 4128
Yearly for 4 years is 4 times
700×4 = 2800
Every month for 4 years is 48 times
100×48 = 4800
Minimum expenditure:
4128 + 2800 + 4800 = $11,728
Step-by-step explanation:
let us give all the quantities in the problem variable names.
x= amount in utility stock
y = amount in electronics stock
c = amount in bond
“The total amount of $200,000 need not be fully invested at any one time.”
becomes
x + y + c ≤ 200, 000,
Also
“The amount invested in the stocks cannot be more than half the total amount invested”
a + b ≤1/2 (total amount invested),
=1/2(x + y + c).
(x+y-c)/2≤0
“The amount invested in the utility stock cannot exceed $40,000”
a ≤ 40, 000
“The amount invested in the bond must be at least $70,000”
c ≥ 70, 000
Putting this all together, our linear optimization problem is:
Maximize z = 1.09x + 1.04y + 1.05c
subject to
x+ y+ c ≤ 200, 000
x/2 +y/2 -c/2 ≤ 0
≤ 40, 000,
c ≥ 70, 000
a ≥ 0, b ≥ 0, c ≥ 0.
Answer: The last option, <u>28</u> minutes
Step-by-step explanation:
See attached for my work. <em>If you are color blind let me know, I color-coded where I "pulled" numbers from.</em>
To answer this problem, we can look at the graph. We need to see for how long the black line is above the blue.
What are the answer choices