Calculus-Maxima and Mininma

Discussion in 'Homework Help' started by Breeze, Sep 27, 2015.

  1. Breeze

    Thread Starter New Member

    Jun 30, 2015
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    Good day to all of you Honorable Engineers,i am a bit lost with how to attempt this calculus problem,it goes like this "A mining company has 10 000 meters of fencing available.It wants to use the fencing to enclose a rectangular field.One side of the field is bordered by a river .If no fencing is placed on the side next to the river,what is the largest area that can be enclosed?

    i assumed my width to be 'x',then divided the field into 3 parts to get the length as a function of width and got this expression: Length=(10 000-3x)/2,but this is where i get lost because the other side of the field that borders the river will not be fenced ,so how can i then derive my equation without the other side being fenced?
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Draw a sketch of the rectangular field. Because it is rectangular, what do you know about the relationships between the lengths of the four sides.

    Taking into account that one side is bounded by the river, what is the perimeter of the field as a function of the variables that describe the lengths of the sides?

    What is the area of the field as a function of the variables that describe the lengths of the sides?

    What constraint does the amount of finding available place on the relationship between the variables that describe the lengths of the sides?

    What is the area of the field as a function of the amount of fencing available and the length of the side bounded by the river?

    How do you find the length of the side that is bounded by the river that maximum the areas enclosed?
     
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  3. Breeze

    Thread Starter New Member

    Jun 30, 2015
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    thanks will look at it that way and see if i can come up with a proper expression
     
  4. Russmax

    Member

    Sep 3, 2015
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    Yes, Breeze, your way sets the ratio of W & L before you begin. The point of your calculations will be to find the W/L ratio that yields maximum area for a given (partial) perimeter, so you'd have to be a lucky person to get the right answer.

    This is really the same problem as if you were fencing all sides of the field, you just have to make an adjustment to your perimeter equation.
     
  5. Breeze

    Thread Starter New Member

    Jun 30, 2015
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  6. Breeze

    Thread Starter New Member

    Jun 30, 2015
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    Very interesting comment,m not short of luck on this platform ....
     
  7. WBahn

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    Mar 31, 2012
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    There's no luck needed, just be methodical. Take it step by step.

    Step 1: Make a sketch of the problem, labeling the width and length of the fence.
     
  8. Breeze

    Thread Starter New Member

    Jun 30, 2015
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    Good morning Senior Engineers-I want to express my gratitude at all your hints you gave me,i did manage to solve it-I got the equation to be (2w+L=10 000) then i solved for L in this equation and substituted that result in the equation for Area..
     
  9. WBahn

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    Mar 31, 2012
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    Glad to hear you got it!
     
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  10. Breeze

    Thread Starter New Member

    Jun 30, 2015
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    Thanks a lot once more,your hints really set me in the right direction.
     
  11. RBR1317

    Active Member

    Nov 13, 2010
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    So what is the largest area that can be enclosed? How do you know that it is the largest?
     
  12. WBahn

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    Mar 31, 2012
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    Reading between the lines, I think this part wasn't the problem for him. He was having problems seeing how to reduce the information to a single function of one variable that could then be optimized.
     
  13. Breeze

    Thread Starter New Member

    Jun 30, 2015
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    The largest Area is got by equating the first derivative (f'x)=0 to get a critical or turning point because we get the local maxima or minimum when the curve is equal to zero.
     
  14. WBahn

    Moderator

    Mar 31, 2012
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    Correction: local maxima or minima when the slope of the curve is equal to zero.
     
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  15. Breeze

    Thread Starter New Member

    Jun 30, 2015
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    Thanks a lot.
     
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