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.

Which one of the following statements is true? Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is correct!*

*Choice (c) is incorrect*

*Choice (d) is incorrect*

Which of the following are the global maximum and global minimum values of the above function. Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is correct!*

Critical points are at $x=-\frac{9}{5}=-1.8$ where ${P}^{\prime}\left(x\right)=0$ and $x=-3$ where ${P}^{\prime}$ is undefined.

Compare the values of the function at the critical points and the endpoints.

$P\left(-1.8\right)=-6.098\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}P\left(-3\right)=0\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}P\left(-4\right)=-12$ and $P\left(-1\right)=-4.762\phantom{\rule{0.3em}{0ex}}.$

So the global minimum is at (-4,-12) and the global maximum is at (-3, 0).

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. (Zero or more options can be correct)

*There is at least one mistake.*

For example, choice (a) should be True.

$\frac{{d}^{2}y}{d{x}^{2}}=12{x}^{2}-12x=12x\left(x-1\right)$

There are stationary points at $x=0$ and $x=\frac{3}{2}\phantom{\rule{0.3em}{0ex}}.$

${\left.\frac{{d}^{2}y}{d{x}^{2}}\right|}_{x=\frac{3}{2}}>0$ so there is a local minimum at $x=\frac{3}{2}\phantom{\rule{0.3em}{0ex}}.$

${\left.\frac{{d}^{2}y}{d{x}^{2}}\right|}_{x=0}=0$ so we cannot decide the nature of the stationary point.

Draw a sign diagram

So there is a stationary point of inflection at $\left(0,0\right)\phantom{\rule{0.3em}{0ex}}.$

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 $\left(\frac{3}{2},-\frac{27}{16}\right)\phantom{\rule{0.3em}{0ex}},$

a stationary point of inflection at $\left(0,0\right)$ and a local maximum at (-1,3).

*There is at least one mistake.*

For example, choice (b) should be False.

*There is at least one mistake.*

For example, choice (c) should be False.

*There is at least one mistake.*

For example, choice (d) should be False.

*There is at least one mistake.*

For example, choice (e) should be True.

*There is at least one mistake.*

For example, choice (f) should be True.

*Correct!*

*True*$\frac{dy}{dx}=4{x}^{3}-6{x}^{2}=2{x}^{2}\left(2x-3\right)$

$\frac{{d}^{2}y}{d{x}^{2}}=12{x}^{2}-12x=12x\left(x-1\right)$

There are stationary points at $x=0$ and $x=\frac{3}{2}\phantom{\rule{0.3em}{0ex}}.$

${\left.\frac{{d}^{2}y}{d{x}^{2}}\right|}_{x=\frac{3}{2}}>0$ so there is a local minimum at $x=\frac{3}{2}\phantom{\rule{0.3em}{0ex}}.$

${\left.\frac{{d}^{2}y}{d{x}^{2}}\right|}_{x=0}=0$ so we cannot decide the nature of the stationary point.

Draw a sign diagram

So there is a stationary point of inflection at $\left(0,0\right)\phantom{\rule{0.3em}{0ex}}.$

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 $\left(\frac{3}{2},-\frac{27}{16}\right)\phantom{\rule{0.3em}{0ex}},$

a stationary point of inflection at $\left(0,0\right)$ and a local maximum at (-1,3).*False*This is a global minimum not a local maximum.*False*This is a stationary point of inflection, not a local minimum.*False*This is a local maximum, not a global maximum.*True*See the explanation above.*True*See the explanation above.

Which of the following gives the correct maximum concentration and the time at which it occurs, to 4 decimal places? Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

Remember to use the product rule and that the derivative of ${e}^{-1.1t}$ is $-1.1{e}^{-1.1t}\phantom{\rule{0.3em}{0ex}}.$

*Choice (c) is correct!*

Since ${x}^{\prime}\left(t\right)=0$ when $t=\frac{1}{1.1}=0.9091$ 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.)

So there is a local maximum at $\left(0.9091,0.1003\right)\phantom{\rule{0.3em}{0ex}}.$

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.

*Choice (d) is incorrect*

Differentiate $x\left(t\right)\phantom{\rule{0.3em}{0ex}},$ using the product rule, and find the critical values.