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Math Help - Calculus Optimization Word Prob

  1. #1
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    Calculus Optimization Word Prob

    Hi, I have 2 questions on these words problems. I really am not sure where to start these.

    1. An advertisement consists of a rectangular printed region with margins of 2 inches each at the top and bottom and 1 inch at each side. If the area of the printed region is to be 98 in^2 , find the overall dimensions if the total area of the advertisement is to be a minimum.

    2. A rancher is going to build a 3-sided enclosure with a divider down the middle. The cost per foot of the 3 side walls are $6/ft, with the single back wall being $10/ft. The area enclosed will be 180 ft^2. What dimensions would minimize the cost?
    The pen looks like this.
    ____
    | | |

    for this one I had an idea of setting 180ft^2 = 3y+x but am not sure if this is the right way to start.
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  2. #2
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    Quote Originally Posted by drewms64 View Post
    1. An advertisement consists of a rectangular printed region with margins of 2 inches each at the top and bottom and 1 inch at each side. If the area of the printed region is to be 98 in^2 , find the overall dimensions if the total area of the advertisement is to be a minimum.
    Let the length and bredth of the ad be l and b, then:

    (l-4)(b-2)=98

    rearranging:

    <br />
l=\frac{98}{b-2}+4<br />

    Then the total area of the ad is:

    <br />
A=l.b=\left(\frac{98}{b-2}+4 \right).b<br />

    So now you just need to find the b which minimises A

    RonL
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  3. #3
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    Quote Originally Posted by drewms64 View Post
    2. A rancher is going to build a 3-sided enclosure with a divider down the middle. The cost per foot of the 3 side walls are $6/ft, with the single back wall being $10/ft. The area enclosed will be 180 ft^2. What dimensions would minimize the cost?
    The pen looks like this.
    ____
    | | |

    for this one I had an idea of setting 180ft^2 = 3y+x but am not sure if this is the right way to start.
    (all dimensions in ft and areas in sq ft)

    The area:

    <br />
A=s \times b=180<br />

    where s is the length of the side walls, and b is the length of the back wall.

    The cost is:

    <br />
C=b \times 10+ 3 \times s \times b<br />

    So you now eliminate one of b or s from the expression for C using the area constraint, then minimise C.

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    Let the length and bredth of the ad be l and b, then:

    (l-4)(b-2)=98

    rearranging:

    <br />
l=\frac{98}{b-2}+4<br />

    Then the total area of the ad is:

    <br />
A=l.b=\left(\frac{98}{b-2}+4 \right).<br />

    So now you just need to find the b which minimises A

    RonL
    Thanks for the advice on both of the questions, for this one would it then end up being,
    x=b y=l
    <br />
A(y)= \left(\frac{98}{y-4}+2 \right)y<br />

    <br />
= \left(\frac{98y}{y-4}+2y \right)<br />

    <br />
A'(y)= \left(\frac{(y-4)(98)-(98y)(1)}{(y-4)^2)}+2 \right)<br />

    <br />
= \left(\frac{-392}{(y-4)^2}+2 \right) = 0<br />

    not sure how to get [tex] to work for this one, but

    -392(y-4)^-2 = -2
    (y-4)^-2 = 196
    take the -2 root of each side
    y-4 = - 14
    y = -10

    then when I find x I get -5

    I dont think these should be negative though?
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  5. #5
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    Pay attention to square root

    Quote Originally Posted by drewms64 View Post
    Thanks for the advice on both of the questions, for this one would it then end up being,
    x=b y=l
    <br />
A(y)= \left(\frac{98}{y-4}+2 \right)y<br />

    <br />
= \left(\frac{98y}{y-4}+2y \right)<br />

    <br />
A'(y)= \left(\frac{(y-4)(98)-(98y)(1)}{(y-4)^2)}+2 \right)<br />

    <br />
= \left(\frac{-392}{(y-4)^2}+2 \right) = 0<br />

    not sure how to get [tex] to work for this one, but

    -392(y-4)^-2 = -2
    (y-4)^-2 = 196
    take the -2 root of each side
    y-4 = - 14
    y = -10

    then when I find x I get -5

    I dont think these should be negative though?
    There is not such thing as -2 root. You should apply the cross product rule:a/b=c/d is equivalent to a*d=b*c.
    therefore, (y-4)^-2=196 becomes (y-4)^2=1/196 and now apply the square root to find
    y-4=1/14 or -1/14
    So, your solutions are y=4+1/14 and y=4-1/14, both positive.
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  6. #6
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    Quote Originally Posted by alinailiescu View Post
    There is not such thing as -2 root. You should apply the cross product rule:a/b=c/d is equivalent to a*d=b*c.
    therefore, (y-4)^-2=196 becomes (y-4)^2=1/196 and now apply the square root to find
    y-4=1/14 or -1/14
    So, your solutions are y=4+1/14 and y=4-1/14, both positive.
    By doing this I got y= 57/14 and by plugging that into the x = (98/(y-4))+2 I get 1372. That seems wrong for the area being 98in^2
    Last edited by drewms64; December 5th 2006 at 07:09 PM.
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  7. #7
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    Correction

    Quote Originally Posted by drewms64 View Post
    By doing this I got y= 57/14 and by plugging that into the x = (98/(y-4))+2 I get 1372. That seems wrong for the area being 98in^2
    Sorry I did not read all of your solution.Your mistake was a little bit sooner.
    -392(y-4)^-2=-2
    (y-4)^-2=-2/-392
    (y-4)^-2=1/196
    (y-4)^2=196
    or y-4=14 or -14
    so, y=18
    x=9
    The dimmension of the printed area being 14, respectively 7 give as the required 98 in^2
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