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Math Help - area maximize problem

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    area maximize problem

    a farmer with 750 feet of fencing wants to enclose a rectangular area and then divide into four pens with fencing parallel to the one side of the rectangle.
    What is the largest possible total area of the four pens?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by gumi View Post
    a farmer with 750 feet of fencing wants to enclose a rectangular area and then divide into four pens with fencing parallel to the one side of the rectangle.
    What is the largest possible total area of the four pens?
    did you draw a diagram? draw one and label all the sides. the perimeter of your figure must be 750. you want to maximize the area of your figure. please do not over-label, it will make your problem harder. that is, label the outside lengths and widths respectively of the fence with ONE variable. you should only have two variables here
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    Quote Originally Posted by gumi View Post
    a farmer with 750 feet of fencing wants to enclose a rectangular area and then divide into four pens with fencing parallel to the one side of the rectangle.
    What is the largest possible total area of the four pens?
    Let l denote the length of the rectangle and w the width.

    The the area is calculated by: ...... a = l \cdot w

    The fence must be divided into one part with the length l
    and 5 parts with the length w:...... l + 5w = 750 ~\implies~l=750-5w

    Plug in the term of l into the first equation:...... a(w) = w \cdot (750-5w)
    Calculate the first derivative: ...... a'(w)= 750 - 10w and solve for w:

    750-10w=0~\iff~w=75 That means l = 750-5\cdot 75 = 375
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    can you tell me why isn't it 2l+5w=750? please explain...i'm lost
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  5. #5
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    Quote Originally Posted by gumi View Post
    can you tell me why isn't it 2l+5w=750? please explain...i'm lost
    Sorry, you are right. I have understood something you didn't write.

    I'm going to repeat the calculations:

    Let l denote the length of the rectangle and w the width.

    The the area is calculated by: ...... a = l \cdot w

    The fence must be divided into two parts with the length l
    and 5 parts with the length w:...... 2l + 5w = 750 ~\implies~l=\frac{750-5w}2

    Plug in the term of l into the first equation:...... a(w) = w \cdot \left(\frac{750-5w}2\right)
    Calculate the first derivative: ...... a'(w)= \frac{750}2 - 5w and solve for w:

    750-10w=0~\iff~w=75 That means l = \frac{750-5 \cdot 75}2 = 187.5
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