 ## MATH1111 Quizzes

Applications: Modelling Quiz
Web resources available Questions

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.

Which of the following formula describes the function to be differentiated to find two positive numbers whose sum is 20 and whose product, $P\phantom{\rule{0.3em}{0ex}},$ is as large as possible. Exactly one option must be correct)
 a) $P=xy\phantom{\rule{0.3em}{0ex}},$ where $x$ and $y$ are two positive numbers less than 20. b) $P=x\left(x-20\right)\phantom{\rule{0.3em}{0ex}},$ where $x$ is a positive numbers less than 20. c) $P\left(x\right)=x\left(20-x\right)\phantom{\rule{0.3em}{0ex}},$ where $0 d) $P=x\left(x+y\right)$ where $x$ and $y$ are two positive numbers less than 20.

Choice (a) is incorrect
Try again, you need to write $P$ in terms of $x$ or $y$ only.
Choice (b) is incorrect
Try again, you are on the right track but this will always give you a negative product.
Choice (c) is correct!
Let $x$ and $y$ be the two positive numbers. Since the sum is 20 we have $x+y=20$ and $y=20-x\phantom{\rule{0.3em}{0ex}}.$
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
Choice (d) is incorrect
Try again, you don’t seem to have read the question correctly.
An open tank is to be constructed with a square base of side $x$ metres with four rectangular sides.
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)
 a) $h=\sqrt{\frac{108}{x}}$ b) $h=\frac{108}{{x}^{2}}$ c) $h=\frac{108}{x}$ d) None of the above, there is not enough information available.

Choice (a) is incorrect
Try again, you seem to have said that $V=x{h}^{2}$ which is not correct.
Choice (b) is correct!
Since the tank has a square base the volume, 108 cubic metres is ${x}^{2}h\phantom{\rule{0.3em}{0ex}},$ the area of the base times the height.
Hence ${x}^{2}h=108\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}h=\frac{108}{{x}^{2}}\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Try again, you seem to have said that $V=xh$ which is not correct.
Choice (d) is incorrect
Try again, we know that $V={x}^{2}h=108\phantom{\rule{0.3em}{0ex}}.$
An open tank is to be constructed with a square base of side $x$ metres with four rectangular sides.
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)
 a) 108 square metres. b) 162 cubic metres. c) 6 square metres. d) 114.3 square metres.

Choice (a) is correct!
The surface area is $A\left(x\right)=\phantom{\rule{0.3em}{0ex}}$area base + area of the 4 sides.
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 $A=36+4×6×3=108$ square metres.
It is unusual for the volume and surface areas to have the same numerical value.
Choice (b) is incorrect
Try again. Recall that the surface area $A={x}^{2}+4xh$ and we need to do several things
• 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
Try again, $x=6\phantom{\rule{0.3em}{0ex}}.$ You may have stopped the problem too soon.
Choice (d) is incorrect
Try again, you may have differentiated the surface area incorrectly.
A magazine advertisement is to contain 50 cm${}^{2}$ of lettering with clear margins of 4 cm each at the top and bottom and 2 cm at each side.
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)
 a) $h=9\phantom{\rule{1em}{0ex}}cm$ b) $h=7.45\phantom{\rule{1em}{0ex}}cm$ c) $h=18\phantom{\rule{1em}{0ex}}cm$ d) $h=10\phantom{\rule{1em}{0ex}}cm$

Choice (a) is incorrect
Try again, this is the optimum width of the advertisement.
Choice (b) is incorrect
Try again, you may not have used the correct formula for the area.
Choice (c) is correct!
The area of the advertisement $A=\left(x+8\right)\left(y+4\right)$ where $x$ is the height of the lettering and $y$ is the width of the lettering. Since $xy=50⇒y=\frac{50}{x}\phantom{\rule{0.3em}{0ex}}.$
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
Try again, this is the height of the lettering of the advertisement.