This quiz tests the work covered in Lecture 19 and corresponds to Section 4.5 of the
textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et
al.).

There are more web quizzes at Wiley, select Section 5.

There is an applet at http://archives.math.utk.edu/visual.calculus/3/applications.2/index.html that explains Example 3, the ladder problem from the textbook. The explanation solves the equation in a much more complex manner than the text but note that it gives the same answer. There are some other applets that you may wish to check out at http://archives.math.utk.edu/utk.calculus/4.6/index.html.

There are some worked examples at http://tutorial.math.lamar.edu/AllBrowsers/2413/Optimization.asp and http://www.math.ucdavis.edu/ kouba/CalcOneDIRECTORY/maxmindirectory/MaxMin.html.

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is correct!*

This gives $P=xy=x\left(20-x\right)\phantom{\rule{0.3em}{0ex}}.$ Both numbers must be greater than zero so they must also be less than 20. Hence $0<x<20\phantom{\rule{0.3em}{0ex}}.$

*Choice (d) is incorrect*

The tank is to have a capacity of 108 cubic metres.

Which of the following gives the correct height, $h\phantom{\rule{0.3em}{0ex}},$ of the tank in terms of $x\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is correct!*

Hence ${x}^{2}h=108\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}h=\frac{108}{{x}^{2}}\phantom{\rule{0.3em}{0ex}}.$

*Choice (c) is incorrect*

*Choice (d) is incorrect*

The tank is to have a capacity of 108 cubic metres.

Which of the following is the least amount of sheet metal from which the tank can be made? Exactly one option must be correct)

*Choice (a) is correct!*

So $A\left(x\right)={x}^{2}+4xh\phantom{\rule{0.3em}{0ex}}.$ Using the previous question we see that $A\left(x\right)={x}^{2}+\frac{432}{x}$

${A}^{\prime}\left(x\right)=2x-\frac{432}{{x}^{2}}=\frac{2{x}^{3}-432}{{x}^{2}}$ and ${A}^{\prime}\left(x\right)=0$ when $2{x}^{3}-432=0$

Hence $x=6$ metres and $h=3$ metres. Hence $A=36+4\times 6\times 3=108$ square metres.

It is unusual for the volume and surface areas to have the same numerical value.

*Choice (b) is incorrect*

- write our formula in terms of $x$
- differentiate the new formula
- find the critical points
- evaluate the function at the critical points and endpoints to determine the global maximum and global minimum.

*Choice (c) is incorrect*

*Choice (d) is incorrect*

Find the dimensions when the total area of the advertisement is a minimum.

Which of the following is the optimum height, $h\phantom{\rule{0.3em}{0ex}},$ of the advertisement. Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is correct!*

Hence $A\left(x\right)=\left(x+8\right)\left(\frac{50}{x}+4\right)=50+4x+\frac{400}{x}+32=4x+\frac{400}{x}+82$

and ${A}^{\prime}\left(x\right)=4-\frac{400}{{x}^{2}}=\frac{4\left({x}^{2}-100\right)}{{x}^{2}}\phantom{\rule{0.3em}{0ex}}.$

If we constructed a sign diagram we would see there is a maximum at $x=-10$ (which doesn’t make physical sense)

and a minimum at $x=10\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{1em}{0ex}}$ So the height if the text is 10 cm.

Hence the height of the advertisement is $h=10+8=18\phantom{\rule{1em}{0ex}}cm\phantom{\rule{0.3em}{0ex}}.$

*Choice (d) is incorrect*