Question

1.

You are given the graph of a function f defined on the interval (−1, ∞). Find the absolute maximum and absolute minimum values of f (if they exist) and where they are attained. (If an answer does not exist, enter DNE.)

The x y-coordinate plane is given. A curve, a horizontal dashed line, and a vertical dashed line are graphed.

• A vertical dashed line crosses the x-axis at x = −1.
• A horizontal dashed line crosses the y-axis at y = 1.
• The curve enters the window in the third quadrant just to the right of x = −1, goes up and right becoming less steep, becomes nearly horizontal at the origin, goes up and right becoming more steep, passes through the approximate point (1, 0.5), goes up and right becoming less steep, and exits the window just below y = 1.

absolute maximum

(x, y)=

absolute minimum(x, y)=

2.

Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)

g(x) = −x² + 4x + 9

maximum =

minimum=

3.

We first note that the function

f(x) = −x² + 2x + 8

is continuous and defined on the closed interval

[3, 6].

Recall that to find the absolute extrema of the given function we must first find any critical numbers of the function that lie in interval

[3, 6].

In other words, we need to find any values of x for which

f '(x) = 0,

or

f '(x)

does not exist.

Therefore, we must first find

f '(x).

f(x)	=	−x2 + 2x + 8
f '(x)	=	−2x +

4.

Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)

f(x) = x² − x − 3 on [0, 3]

maximum=

minimum=

5.

Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)

f(x) = 4x −9/X on [9, 11]

maximum=

minimum=

6.

Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)

f(x) =

1

x2 + 2x + 9

on [−2, 1]

maximum=

minimum=

7.

Average Speed of a Vehicle

The average speed of a vehicle on a stretch of Route 134 between 6 A.M. and 10 A.M. on a typical weekday is approximated by the function

f(t) = 27t − 54

t

+ 64    (0 ≤ t ≤ 4)

where f (t) is measured in miles per hour, and t is measured in hours, with t = 0 corresponding to 6 A.M. At what time of the morning commute is the traffic moving at the slowest rate?

=A.M.

What is the average speed of a vehicle at that time?

=mph

8.

Maximizing Profits

The quantity demanded each month of the Walter Serkin recording of Beethoven's Moonlight Sonata, produced by Phonola Media, is related to the price per compact disc. The equation

p = −0.00054x + 6    (0 ≤ x ≤ 12,000)

where p denotes the unit price in dollars and x is the number of discs demanded, relates the demand to the price. The total monthly cost (in dollars) for pressing and packaging x copies of this classical recording is given by

C(x) = 600 + 2x − 0.00003x²    (0 ≤ x ≤ 20,000).

To maximize its profits, how many copies should Phonola produce each month? Hint: The revenue is

R(x) = px,

and the profit is

P(x) = R(x) − C(x).

(Round your answer to the nearest whole number.)

= discs/month

9.

Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)

g(x) =

1

16

x² − 16

x

on [0, 36]

maximum=

minimum=

10.

Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)

g(t) =

t

t − 2

on [4, 6]

maximum=

minimum=

11.

Enclosing the Largest Area

The owner of the Rancho Grande has 3,044 yd of fencing with which to enclose a rectangular piece of grazing land situated along the straight portion of a river. If fencing is not required along the river, what are the dimensions (in yd) of the largest area he can enclose?

A rectangular piece of land has been enclosed along a straight portion of a river. The enclosure is bordered by the river (a long side), another long side of fence, and two short sides of fence.

shorter side = yd

longer side= yd

What is this area (in yd²)?

= yd²

^12.

Packaging

By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. If the cardboard is 16 in. long and 6 in. wide, find the dimensions (in inches) of the box that will yield the maximum volume. (Round your answers to two decimal places if necessary.)

smallest value = in

= in

largest value = in

13.

Minimizing Packaging Costs

A rectangular box is to have a square base and a volume of 72 ft³. If the material for the base costs $0.49/ft², the material for the sides costs $0.12/ft², and the material for the top costs $0.15/ft², determine the dimensions (in ft) of the box that can be constructed at minimum cost. (Refer to the figure below.)

A closed rectangular box has a length of x, a width of x, and a height of y.

x= ft

y= ft

14.

Charter Revenue

The owner of a luxury motor yacht that sails among the 4000 Greek islands charges $568/person/day if exactly 20 people sign up for the cruise. However, if more than 20 people sign up (up to the maximum capacity of 100) for the cruise, then each fare is reduced by $4 for each additional passenger.

Assuming at least 20 people sign up for the cruise, determine how many passengers will result in the maximum revenue for the owner of the yacht.

= passengers

What is the maximum revenue?

$ =

What would be the fare/passenger in this case? (Round your answer to the nearest dollar.)

= dollars per passenger

15.

Minimizing Packaging Costs

If an open box has a square base and a volume of 103 in.³ and is constructed from a tin sheet, find the dimensions of the box, assuming a minimum amount of material is used in its construction. (Round your answers to two decimal places.)

height = in

length = in

width= in

Answer 1

1

As per mentioned property, curve should look like above. There is no maximum and minimum . DNE.

2>

differentiating both side to get the critical points:

, since 2nd derivative is negative at critical point , x = 2 is maximum.

maximum = at x = 2, g(2) = 13

minimum= DNE

3.

is continuous and defined on the closed interval [3, 6].

The critical values lies in between the required interval.

, this is the point of maximum .

maximum = at x = 1, f(1) = 9

minimum= DNE

4>

, this critical value lies in the given interval.

, since 2nd derivative is positive, this critical point gives minimum value.

maximum= DNE

minimum= -3.25