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Expressing functions [word problems]

  1. Sep 18, 2011 #1
    Could somebody help me understand the process of solving these problems because I don't even know where to start... Also, could someone provide a link to a site where I can practice more of these type of problems.
    1) The problem statement, all variables and given/known data:
    A rectangular package to be sent by a postal service can have a maximum combined length and girth (perimeter of a cross section) of 108". Express the volume of such a package as a function of x.

    2) The problem statement, all variables and given/known data:
    A Palladian window is a rectangle with a semicircle on top, as shown below. Suppose the perimeter of a particular Palladian window is 28 ft. Express the total area of the window as a function of x.
    [PLAIN]http://rghost.ru/22034911/image.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 18, 2011 #2

    symbolipoint

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    Some guidance for #2, the palladian window problem:

    Label any other part of the diagram that you need. Create expressions and equations or relations for any quantities that you can identify. Use the expressions, variables, relations, to setup your equation or relation or system of relations to solve for what variable you need.

    Let A be total area. x is both the diameter of the half circle and one length of the rectangle part; y is the other length of the rectangle part.

    Total area, A: [itex]A=\frac{1}{2}\pi\frac{x^2}{2^2}+xy[/itex]

    Length of half-circle is: [itex]\frac{1}{2}2\pi\frac{x}{2}[/itex]

    You want the perimeter to be 28 feet. Your turn... What is an equation for this perimeter? How will you find a substitution for 'y'? You want ultimately A as a function of x, without further relying on y.
     
  4. Sep 18, 2011 #3
    symbolipoint i have no idea where are you getting the formulas and/or how are you setting up the equation the way you did?
     
  5. Sep 18, 2011 #4

    symbolipoint

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    For now, I will leave my posting the way I wrote it, but I will recheck my work. You could try to derive the equations yourself based on the known formulas and maybe find any mistakes I might have made. (My work has less chance of 'error' when I do all of it with pencil on paper).

    EDIT: Just done checking my work, on paper. I found no mistakes.
    Problem #2 has about all the parts labeled which are necessary. The half circle has radius of [itex]\frac{x}{2}[/itex]. What is the length of just this half circle?
    You should be able to write an equation for Area and an equation for Perimeter. You will need to make a substitution to transform Area as a function of only x.
     
    Last edited: Sep 18, 2011
  6. Sep 18, 2011 #5
    symbolipoint problem is i dont understand where are you getting that
    [tex]Radius = \frac{\pi }{2}[/tex].. the questions you're asking me im kinda of lost :)
     
  7. Sep 18, 2011 #6

    symbolipoint

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    The radius of the half-circle is [itex]\frac{x}{2}[/itex]. The circumerence of the whole circle would be [itex]2\pi\frac{x}{2}[/itex]
     
  8. Sep 18, 2011 #7
    AARMA
    It would help if we knew what grade level you are at.
    Teaching a 6th grader one assumes they only know Arithmetic
    But a 10th grader should have had Algebra and Geometry for several years.

    and

    Where did you get these problems ?
    They are in general not solvable because no information is given
    for the relationship of x to the other unknown sides.
    Both problems ask for the answer as a function of x only.
    You can answer #2 only if you know y = f(x)
    You can answer #1 only if you know y = f(x) and z = f(x)
     
    Last edited: Sep 18, 2011
  9. Sep 18, 2011 #8
    paulfr, well, in my personal opinion, public education in US is very weak and does not provide the very needs (the basics) for higher education.. As my college professor says forget everything that you have been taught in high school; high school is like 12 years of horror, but this is college now... I agree with him.. I mean our Precalculus teacher didn't even want to show us more examples and attempted to cram everything known in Mathematics in just half a year... :)

    as far as problem goes: this is how i would explain this problem to someone... we must write the area as Area = exes; thus we're not looking for a particular ANSWER we're looking how well we can manipulate equations and make subsititutions....
    Formulas to consider:
    Area of a circle = [tex]\pi r^{2}[/tex] ---> now essentially, we have half a circle [semicircle] thus we must divide by 2
    Circumference of a circle = [tex]2\pi r[/tex] ---> now essentially, we have a half circle [semicircle] thus we must divide by 2 on both sides
    -------------------------------
    Because we are given the quantity of the perimeter we first must solve for it:
    [tex]x+2y+(\pi\cdot r)[/tex]
    [tex]Radius = \frac{x}{2}[/tex] ---> this is so because we have a diameter of our circle (x) in order to have a radius we take x and divide it by 2
    [tex]x+2y+\frac{x}{2}\pi[/tex]
    [tex]2y + x(1+\frac{\pi }{2}) = 28[/tex] ---> we factor out the x and plug in the 28 ft as our GIVEN perimeter
    [tex]2y + x(1+\frac{\pi }{2})- x(1+\frac{\pi }{2})= 28 ft - x(1+\frac{\pi }{2})[/tex] ---> here, what we're doing is, we're taking the factored out terms and subtracting them from both sides (in other words moving/isolating the y)
    [tex]\frac{2y}{2} = \frac{28}{2} - \frac{x}{2}(1 + \frac{\pi}{2})[/tex]
    [tex]y = 14 - \frac{x}{2}(1 +\frac{\pi }{2})[/tex] ---> we divide everything [on both sides] by 2 in order to have 1 y
    So, now solving for the area
    [tex]xy+ \frac{\pi r^{2}}{2}[/tex]
    [tex]xy + \frac{1}{2}\pi (\frac{x}{2})^{2}[/tex] ---> here, we're taking the radius and for radius we're using x/2 again and taking the formula for half of area of a circle
    [tex]x(14-\frac{x}{2}(1+\frac{\pi }{2}))+\frac{1}{2}\pi \frac{x^{2}}{4}[/tex] ---> now we know that y = ... thus we plug in the y into the xy
    [tex]x(14-\frac{x}{2}(1+\frac{\pi }{2}))+ \frac{x^{2}\pi }{8}[/tex]
    [tex]14x - \frac{x^{2}}{2} - \frac{x^{2}\pi }{4} + \frac{x\pi }{8}[/tex]
    [tex]14x - \frac{x^{2}}{2} - \frac{x^{2}\pi }{8} = Area[/tex]
     
    Last edited: Sep 18, 2011
  10. Sep 18, 2011 #9
    Well I stand corrected.
    That is a nice solution to #2
    Well done

    But #1 is still unsolvable as the problem does not even state what x is
    much less the other two sides of the rectangular prism.

    Most of the problems of US education is the lowering of standards to keep grades high [a form of grade inflation] with declining levels of work done by the student.
    The easy way to high grades feels good to parents, students, administrators and teachers but the long term consequences are not . And students are in for a big shock when they get to University and the demands of PhD professors.
     
  11. Sep 20, 2011 #10
    [PLAIN]http://rghost.ru/22429021/image.png [Broken]
    paulfr, can you help me with this one?
    here's the picture [quickly made it in Photoshop (dont critique :))]
    the part i dont understand is:
    108 = 2*x + 2*y ---> why are we doing this?

    And please, can you give some reference where i can find more of these problems? I need to PRACTICE
    thanks in advance....
     
    Last edited by a moderator: May 5, 2017
  12. Sep 20, 2011 #11
    This problem is quite unrepresentative of the word problems of high school.
    Go here to PurpleMath to see a whole section devoted to them
    http://www.purplemath.com/modules/index.htm

    Begin with the "Translation" subtopic.
    And note how important it is to learn how to make and use charts or grids
    to organize the information. This makes finding the equation to be solved
    much easier.
    I suggest doing the age, area, distance and quadratic subtopics first before
    doing the other subtopics.

    As for your length + girth problem, the drawing looks wrong as the long length
    and the width both are labeled "x" but are clearly different in length.
    Ignoring this we have .........
    2x + 2y equals the formula for Girth, the perimeter.
    The L+G value would yield 3x+2y = 108 so y=54 - (3/2) x
    Then Volume would be ... V = x * x * y = x * x * (54 - (3/2) x)
    So V = f(x) = [54x^2] - [(3/2) x^3]

    Cheers
     
  13. Sep 20, 2011 #12

    symbolipoint

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    AARMA,

    Although you did fine work in #2, we still say #1 is incompletely described. What is the exact problem description for #1, and could you represent a figure in the book for this if it is shown? Is there maybe more information to relate width to girth, or the girth, may be like based on lengths, x1 and x2 ? In such case, perimeter is
    108=2x1 + 2x2.
    Your formula for volume, V, would be V=(x1)(x2)(y), assuming the other dimension takes assigned variable, y. Now we can use the length & girth relation to eliminate one or the other of the x's, but that seems the best we can do until we have more information given.

    You can find several exercises such as you asked from any good College Algebra book and from any good Pre-Calculus book.
     
  14. Sep 20, 2011 #13
    paulfr + symbolipoint, these word problems are not from the book but rather were given to us from the professor as a worksheet. the diagrams were drawn by him... For the rectangular package diagram i, myself, got confused with the variable correspondence for each side... Now, however, i understand your point
    [tex]Volume = x^{2}*y[/tex]
    [tex]108 = 2x + 2y[/tex] <---- this comes from the fact that girth is a cross section [a square]? you add 4 sides
    so then in the problem it asks for girth + length... shouldn't it be 2x + 2y and plus another variable for the length?
    [tex]54 - x =y[/tex]
    [tex]V = x^{2}(54-x)[/tex] <--- this is the answer we got in class
     
  15. Sep 21, 2011 #14

    symbolipoint

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    This article, helps explain how question #1 works:
    http://en.wikipedia.org/wiki/Girth

    Length plus girth, meaning y+4x<=108, assuming the cross section is based on a square, each side of this square being x.
     
  16. Sep 21, 2011 #15

    symbolipoint

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    Your question #1 about the rectangular box with the girth plus length condition is nearly identical to the Precalculus book by Larson, Hostetler, Edwards from year 1993, except that in that book problem, a volume value was given as a constant. Section 3.4, Real Zeros of Polynomial Functions, in the section exercises. A figure drawing was included; the girth was based on a square cross section of x by x. Combined girth plus length was given as 108 inches.

    Your teacher just gave the same exercise with a minor change, for you to not be given volume, but to find a formula for volume.
     
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