0.08*x+0.03*(29000-x)<span> </span>
Answer:
<u><em>x>2</em></u>
Step-by-step explanation:
<u><em>x3+3x2=7x+6</em></u>
<u><em>x3+3x2−(7x+6)=7x+6−(7x+6)</em></u>
<u><em>x3+3x2−7x−6=0</em></u>
<u><em>(x−2)(x2+5x+3)=0</em></u>
<u><em>x−2=0 or x2+5x+3=0</em></u>
<u><em>x=2 or x=−0.6972243622680054 or x=−4.302775637731995</em></u>
<u><em>x<−4.302776</em></u>
<u><em>−4.302776<x<−0.697224</em></u>
<u><em>−0.697224<x<2</em></u>
<u><em>x>2(Works in original inequality)</em></u>
Answer:
Step-by-step explanation:
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.
