Real World Quadratic Functions Problem #56 Maximum profit. A chain store manager has been told by the main office that daily profit, P , is related to the number of clerks working that day, x ,...

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Answered Same DayDec 22, 2021

Answer To: Real World Quadratic Functions Problem #56 Maximum profit. A chain store manager has been told by...

David answered on Dec 22 2021
138 Votes
Sol: Profit function
2
P(x)=-25x300x,
+
where x is the number of clerks working.
This equation is similar to the equation of
parabola
2
,
yaxbxc
=++
Where
25, 300, and 0,
abc
=-==
The x-intercepts of the parabola can be found by solving
2
-25x300x=0.
+

(
)
2
-25x300x=0 Divide both sides b
y -1.
2
25x-300x=0 Factor the left
side.
25xx-120 Use Zero Factor Pr
operty.
25x=0 or x-12=0 Solve each equa
tion.
x=0 or x=12
+
=
The parabola will cross t
he x-axis at 0 and 12.
This quadratic function has a large
a
value which means the parabola will be narrow. It also has a negative
a
value so the parabola will open downward. This means there will be maximum value of the graph at the vertex which will happen at the x value of
,
2
b
a
-
where
300 and 25
ba
==-
in this case.
What number of clerks will maximize the profit?

(
)
300150
6 6 clerks working will maxi
mize
2
22525

profit.
b
a
-
-
===
-
What is the maximum possible profit when this many clerks are working?

(
)
(
)
(
)
(
)
(
)
(
)
(
)
2
P62563006
P625361800
P69001800
P6900 ...
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