Question

Let the charge per cup of coffee be
x
$
. Then, the number of cups that will be sold is
1600
+
50
×
2.40
−
x
0.05
=
1600
+
(
2400
−
1000
x
)
=
4000
−
1000
x
Thus, the revenue will be
x
(
4000
−
1000
x
)
$
=
1000
x
(
4
−
x
)
$
. In order to maximize revenue, we need to maximize the quantity
x
(
4
−
x
)
. While one can use calculus and equate the first derivative to zero to solve the problem, I personally favor the algebraic approach:
x
(
4
−
x
)
=
−
(
x
2
−
4
x
)
=
4
−
(
x
2
−
4
x
+
4
)
=
4
−
(
x
−
2
)
2
Since a perfect square can never be negative, the largest value
x
(
4
−
x
)
can attain is 4, and this value is attained for
x
=
2
.
At the rate of
2
$
, the student center will sell 2000 cups of coffee, maximizing its revenue at
4000
$
up
160
$
from the revenue it is posting now.

Answer 1

do you can post an ss please about your problem ? maybe more easy understandably

Answer 2

Wait a second.

Answer 3

There's no question here.

Answer 4

It's just stating how they solved it, no question to solve for.