You are here:   Maths & Stats Home / Teaching program / Junior / MATH1111 / Quizzes / Applications

Applications: Optimization Quiz

Last unanswered question  Question  Next unanswered question

Web resources available

This quiz tests the work covered in Lecture 18 and corresponds to Section 4.3 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).
There are more web quizzes at Wiley, select Section 3.

Section 4 of The Mathematics Learning Centre’s booklet on differentiation Introduction to Differential Calculus covers optimization.

There is a good summary of the theory at http://www.math.hmc.edu/calculus/tutorials/secondderiv/.

There are worked examples at http://tutorial.math.lamar.edu/AllBrowsers/2413/AbsExtrema.asp.

Question 1

Suppose f is continuous on the closed interval - 5 ≤ x ≤ 4.
Which one of the following statements is true?

a)
The global maximum and minimum can only occur at either x = - 5 or x = 4.
b)
The global maximum and minimum can only occur at either x = - 5 or x = 4 or critical points of f.
c)
The global maximum and minimum can only occur at either x = 5 or x = - 4 or where f′(x) = 0.
d)
The global maximum and minimum can only occur at either x = 5 or x = - 4 or at critical points of f.

 

Not correct. Choice (a) is false.
Try again, it can have a maximum or minimum at critical points of f as well.
Your answer is correct.
We know from Theorem 4.2 that there is a global maximum and minimum and it occurs at the endpoints or where f′(x) = 0 or where f′ is undefined - the critical points.
Not correct. Choice (c) is false.
Try again, you do not have the correct endpoints, you also need to consider where f′ is undefined.
Not correct. Choice (d) is false.
Try again, you do not have the correct endpoints.

Question 2

Consider the function P(x) = 3x(x+ 3)23 on the interval [- 4,- 1].
Which of the following are the global maximum and global minimum values of the above function.

a)
The global minimum is at (-4,-12) and the global maximum is at (-1, -4.762).
b)
The global minimum is at (-4,-12) and the global maximum is at (-1.8, -6.098).
c)
The global minimum is at (-3,0) and the global minimum is at (-4, 12).
d)
The global minimum is at (-4,-12) and the global maximum is at (-3, 0).

 

Not correct. Choice (a) is false.
Try again, you may not have considered where P′(x) is undefined.
Not correct. Choice (b) is false.
Try again, you must look at the derivative function to find the critical values for P.
Not correct. Choice (c) is false.
Try again, you have not evaluated the function correctly at x = - 4. Note that (- 1)23 = 1.
Your answer is correct.
2 2 1 5x + 913 P ′(x) = 3(x +3)3 + 3x×-(x + 3)- 3 = ------ 3 (x + 3
Critical points are at x = - 9= - 1.8 5 where P′(x) = 0 and x = - 3 where P ′ is undefined.
Compare the values of the function at the critical points and the endpoints.
P (- 1.8) = - 6.098, P(- 3) = 0, P (- 4) = - 12 and P(- 1) = - 4.762.
So the global minimum is at (-4,-12) and the global maximum is at (-3, 0).

Question 3

Consider the function y = x4 - 2x3 for x between -1 and 3.
Find the stationary points and points of inflection and any other points of interest to decide which of the following statements are true.
There may be more than one correct answer.

a)
There is a global maximum at (3,27).
b)
There is a local maximum at (3,- 27). 2 16
c)
There is a local minimum at (0,0).
d)
There is a global maximum at (- 1,3).
e)
There is a global minimum at (3,- 27). 2 16
f)
There is a stationary point of inflection at (0,0).

 

There is at least one mistake.
For example, choice (a) should be true.
dy- 3 2 2 dx = 4x - 6x = 2x (2x- 3)
-d2y = 12x2 - 12x = 12x(x- 1) dx2
There are stationary points at x = 0 and 3 x = 2.
| d2y|| > 0 dx2|x= 32 so there is a local minimum at x = 3. 2
d2y|| dx2|| = 0 x=0 so we cannot decide the nature of the stationary point.
Draw a sign diagram

PIC
So there is a stationary point of inflection at (0,0).
The function is on a restricted domain so we find the endpoints are (-1,3) and (3,27).
Finally there is a global maximum at (3,27), a global minimum at (3,- 27), 2 16
a stationary point of inflection at (0,0) and a local maximum at (-1,3).
There is at least one mistake.
For example, choice (b) should be false.
This is a global minimum not a local maximum.
There is at least one mistake.
For example, choice (c) should be false.
This is a stationary point of inflection, not a local minimum.
There is at least one mistake.
For example, choice (d) should be false.
This is a local maximum, not a global maximum.
There is at least one mistake.
For example, choice (e) should be true.
See the explanation above.
There is at least one mistake.
For example, choice (f) should be true.
See the explanation above.
Your answers are correct
  1. True. dy- 3 2 2 dx = 4x - 6x = 2x (2x- 3)
    -d2y = 12x2 - 12x = 12x(x- 1) dx2
    There are stationary points at x = 0 and 3 x = 2.
    | d2y|| > 0 dx2|x= 32 so there is a local minimum at x = 3. 2
    d2y|| dx2|| = 0 x=0 so we cannot decide the nature of the stationary point.
    Draw a sign diagram

    PIC
    So there is a stationary point of inflection at (0,0).
    The function is on a restricted domain so we find the endpoints are (-1,3) and (3,27).
    Finally there is a global maximum at (3,27), a global minimum at (3,- 27), 2 16
    a stationary point of inflection at (0,0) and a local maximum at (-1,3).
  2. False. This is a global minimum not a local maximum.
  3. False. This is a stationary point of inflection, not a local minimum.
  4. False. This is a local maximum, not a global maximum.
  5. True. See the explanation above.
  6. True. See the explanation above.

Question 4

The concentration of a certain drug in the blood at time t hours after taking the dose is x units, where x(t) = 0.3te-1.1t for the first 3 hours.
Which of the following gives the correct maximum concentration and the time at which it occurs, to 4 decimal places?

a)
At time t = 0.9090 hours the maximum value of 0.1003 units is reached.
b)
At time t = 1 hour the maximum value of 0.0999 units is reached.
c)
At time t = 0.9091 hours the maximum value of 0.1003 units is reached.
d)
At time t = 0 hours the maximum value of 0.3 units is reached.

 

Not correct. Choice (a) is false.
Try again, you have not rounded your value for t correctly.
Not correct. Choice (b) is false.
Try again, t = 1 is not a critical value, you may not have differentiated correctly.
Remember to use the product rule and that the derivative of e-1.1t is - 1.1e-1.1t.
Your answer is correct.
x′(t) = 0.3e-1.1t + 0.3t(- 1.1)e- 1.1t = 0.3e-1.1t(1- 1.1t)
Since x′(t) = 0 when t =-1 = 0.9091 1.1 there is a critical point at (0.9091,0.1003).
Draw a sign diagram to determine the nature of the critical point.
(For complicated functions it is often easier to draw the sign diagram than to find the second derivative.)

PIC
So there is a local maximum at (0.9091,0.1003).
The values at the extrema are (0,0) and (3,0.0332)
so the maximum value is at 0.1003 units at time t = 0.9091 hours.
Not correct. Choice (d) is false.
Try again, x(0) = 0 so this cannot be the correct answer.
Differentiate x(t), using the product rule, and find the critical values.
For questions or comments please contact webmaster@maths.usyd.edu.au