Learning Activity for MATH 115
Applied Math for Business Fall 2013
NAME:
Section:
Answer each of the questions below to the best of your ability. Be sure to answer all word problems
using full sentences that include units for the problem.
Questions
1. Corner Stone Electronics determines that its weekly profit, in dollars, from the production
and sale of x amplifiers is given by
p(x) =
x2
1500
.
− 6x + 10
Find the number of amplifiers, x, for which the total weekly profit is a maximum.
First we find the critical points of p(x) by looking for values of x where p(x) is not defined
or is equal to zero.
p(x) = 1500(x2 − 6x + 10)−1
=⇒ p0 (x) = −1500(x2 − 6x + 10)−2 (2x − 6)
−3000x + 9000
=⇒ p0 (x) =
(x2 − 6x + 10)2
We have critical points when
2
x − 6x + 10 = 0 =⇒ x =
or when
6±
p
√
36 − 4(10)
6 ± −4
=
both complex
2
2
p0 (x) = 0 =⇒ −3000x + 9000 = 0 =⇒ x = 3.
we can classify this point using the first or second derivative test. Checking a value on each
side of three.
p0 (2) = 750 and p0 (4) = −750 =⇒ p(3) is a Rel Max.
The profit of $1500, from producing and selling three amplifiers will be the maximum profit.
2. The total cost and total revenue functions for producing x items are:
c(x) = 5000 + 600x
and
1
r(x) = − x2 + 1000x,
2
where 0 ≤ x ≤ 600. Use these functions to:
• Find the total Profit function p(x).
• Find the number of items, x for which the total profit is a maximum.
The profit function p(x) is found to be:
1
p(x) = r(x) − c(x) = − x2 + 1000x − 5000 − 600x
2
1 2
= − x + 400x − 5000
2
We can find the maximum profit by inspecting the critical points of p(x):
p0 (x) = −x + 400
Note the derivative exists everywhere. Consider p0 (x) = 0
=⇒ x = 400.
Using the first derivative test we see the function is increasing for x values less than 400, and
decreasing for x values after 400. Thus, the profit is maximized at a value $75,000 of when
producing 400 items.
3. A lifeguard needs to rope off a rectangular swimming area in front of IU P Beach, using 180
yards of rope and floats. What dimensions of the rectangle will maximize the area? (Note
shore will be one side of the rectangle.)
(Draw a picture!) Let x be the length of a non-shore side of the rectangle. Here we note
that the total length of the rope used is 180 yards, thus the length of the shoreline side of the
rectangle is 180 − 2x. The area of the swimming space as a function of x is
A(x) = x(180 − 2x) = −2x2 + 180x
Looking for the critical points gives:
A0 (x) = −4x + 180 =⇒ 0 = −4x + 180 =⇒ x = 45.
From the second derivative A00 (x) = −4 we know that function is concave down giving a
relative maximum at x = 45. The non-shoreline side of the swimming space should be 45
yards and the water front should be 90 yards long.
4. Ever Green Gardening is designing a rectangular compost container that will be twice as tall
as it is wide and must hold 18ft3 of composted material. Find the dimensions of the compost
container with minimal surface area (include the bottom and top).
Here we let x be the width. Note that then each side may be represented in terms of x using
the fact that the volume of the container is 18ft3 .
y=
9
x2
The surface area is then given by:
S(x) = 2(x)
9
x2
+ 2(x)(2x) + 2(2x)
=⇒ S(x) =
18
x
+ 4x2 +
2
=⇒ S(x) = 4x +
54
x
36
x
9
x2
where 0 ≤ x ≤ 9. We look at the critical points of the surface area function:
S 0 (x) =
−54
+ 8x
x2
Reject x = 0 as the box would have no dimensions. Consider s0 (x) = 0, then
−
54
+ 8x = 0 =⇒ x = (54/8)1/3 ≈ 1.88988157484
x2
We use the second derivative test to see that this is indeed a minimum for the surface area:
s00 (x) = 104x−3 + 8
s00 (1.88988157484) ≈ 92.11791039
Thus, the surface area function is concave up at the critical point yielding a minimum surface
area of 42.85982 ft2 for the compost container when the width is approximately 1.89 ft.