Results 1 to 3 of 3

Math Help - perimeter and area

  1. #1
    Senior Member furor celtica's Avatar
    Joined
    May 2009
    Posts
    271

    perimeter and area

    the cross-section of an object has the shape of a quarter-circle of radius r adjoining a rectangle ofwidth x and height r as shown in the diagram. A. the perimeter and area of the cross-section are P and A respectively. express each of P and A in terms of r and x, and hence show that A=0.5Pr - r^2 B. taking the perimeter of the cross-section as fixed, find x in terms of r for the case when the area A of the cross section is a maximum, and show that for this value of x A is a maximum and not a minimum. ok the part A. was on problem. but part B. i kinda dont even understand what they are saying, perimeter 'as fixed'? i dont get. i don't understand how to logically combine the given equations, can someone help me wokr it through?
    Attached Thumbnails Attached Thumbnails perimeter and area-p.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    Perimeter P = πr/2 + 2r + 2x.

    Area of cross section A = π*r^2/4 + xr

    Substitute the value of x from the above equation. Find dA/dr and equate it to zero. Hence solve for r.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by furor celtica View Post
    the cross-section of an object has the shape of a quarter-circle of radius r adjoining a rectangle of width x and height r as shown in the diagram.

    (A). the perimeter and area of the cross-section are P and A respectively. express each of P and A in terms of r and x, and hence show that A=0.5Pr - r^2

    (B). taking the perimeter of the cross-section as fixed, find x in terms of r for the case when the area A of the cross section is a maximum, and show that for this value of x A is a maximum and not a minimum.

    ok the part A. was no problem, but part B ? i kinda dont even understand what they are saying, perimeter 'as fixed'? i dont get. i don't understand how to logically combine the given equations, can someone help me work it through?
    If the external perimeter is fixed, then if you increase r, x must decrease.
    This is because P requires both x and r and if you change one,
    you must change the other also, in order to keep P constant.

    P=\frac{2{\pi}r}{4}+2x+2r contains x and r.

    A=\frac{{\pi}r^2}{4}+xr=0.5Pr-r^2=-r^2+kr

    since P is a constant.
    The graph of A versus r is a quadratic.
    It's highest power of r co-efficient is negative.
    Therefore the graph is an inverted U-shape.
    This means it's derivative is zero only at the maximum,
    which is mid-way between the two roots (A=0).

    You may be able to finish up.
    If not, get back on this.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Perimeter and Area
    Posted in the Geometry Forum
    Replies: 3
    Last Post: February 10th 2010, 03:19 PM
  2. Perimeter Area
    Posted in the Geometry Forum
    Replies: 8
    Last Post: October 5th 2008, 07:58 AM
  3. Area from perimeter is it possible?
    Posted in the Geometry Forum
    Replies: 6
    Last Post: July 20th 2008, 05:25 AM
  4. perimeter and area
    Posted in the Geometry Forum
    Replies: 4
    Last Post: April 8th 2008, 06:02 PM
  5. Area and Perimeter
    Posted in the Geometry Forum
    Replies: 4
    Last Post: May 1st 2007, 05:43 PM

Search Tags


/mathhelpforum @mathhelpforum