A cleaning company charges $ 100 for each office it cleans. The fixed monthly cost of $ 480 for the company includes telephone service and the depreciation on cleaning equipment and a van. The variable cost is $ 52 per office and includes labor, gasoline, and cleaning supplies. (See Example 9) a. Write a linear cost function representing the cost C x in $ to the company to clean x offices per month. b. Write a linear revenue function representing the revenue R x in $ for cleaning x offices per month. c. Determine the number of offices to be cleaned per month for the company to break even. d. If 28 offices are cleaned, will the company make money or lose money?
A cleaning company charges $ 100 for each office it cleans. The fixed monthly cost of $ 480 for the company includes telephone service and the depreciation on cleaning equipment and a van. The variable cost is $ 52 per office and includes labor, gasoline, and cleaning supplies. (See Example 9) a. Write a linear cost function representing the cost C x in $ to the company to clean x offices per month. b. Write a linear revenue function representing the revenue R x in $ for cleaning x offices per month. c. Determine the number of offices to be cleaned per month for the company to break even. d. If 28 offices are cleaned, will the company make money or lose money?
A cleaning company charges
$
100
for each office it cleans. The fixed monthly cost of
$
480
for the company includes telephone service and the depreciation on cleaning equipment and a van. The variable cost is
$
52
per office and includes labor, gasoline, and cleaning supplies. (See Example 9)
a. Write a linear cost function representing the cost
C
x
in $
to the company to clean
x
offices per month.
b. Write a linear revenue function representing the revenue
R
x
in $
for cleaning
x
offices per month.
c. Determine the number of offices to be cleaned per month for the company to break even.
d. If 28 offices are cleaned, will the company make money or lose money?
(b) Find the (instantaneous) rate of change of y at x = 5.
In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of
change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the
following limit.
lim
h→0
-
f(x + h) − f(x)
h
The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule
defining f.
f(x + h) = (x + h)² - 5(x+ h)
=
2xh+h2_
x² + 2xh + h² 5✔
-
5
)x - 5h
Step 4
-
The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x).
-
f(x + h) f(x) =
= (x²
x² + 2xh + h² -
])-
=
2x
+ h² - 5h
])x-5h) - (x² - 5x)
=
]) (2x + h - 5)
Macbook Pro
Evaluate the integral using integration by parts.
Sx² cos
(9x) dx
Let f be defined as follows.
y = f(x) = x² - 5x
(a) Find the average rate of change of y with respect to x in the following intervals.
from x = 4 to x = 5
from x = 4 to x = 4.5
from x = 4 to x = 4.1
(b) Find the (instantaneous) rate of change of y at x = 4.
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