You have been asked to make a rectangular enclosure to keep the deer out. A rectangular fence is to be constructed from 84 ft of materials.?

The area enclosed by the fence is given by

A(w) = w(84 - 2w)

where A(w) represents the area enclosed in a rectangle with a width of w ft.

Find the length and width that maximize the area.

Draw a scale diagram of the fencing with the length and width that maximizes area.

1 mark for showing appropriate work to find width

1 mark for correct width

1 mark for showing appropriate work to find length

1 mark for correct length

1 mark for scale diagram

5 Answers

  • 1 month ago
    Favourite answer

    There seems to be some missing information.

    From the equation A(w) = w(84 - 2w)

    it seems one length of the enclosure is a wall not fencing.

    Instead of the perimeter being 2 widths and 2 lengths

    the fencing used for the perimeter is 2 widths and 1 length.

    2w + l = 84

    l = 84 - 2w

    which is the second term in the area equation


    w(84 - 2w) are the factors of the quadratic A = -2w² + w84

    which is a parabola that opens down.

    Therefor the vertex is the maximum

    The maximum is the midpoint of the roots

     w = 0

    84 - 2w = 0 ===> w = 42

    w at vertex

    w = (0 + 42) / 2 = 21 

    width that maximizes area = 21ft <–––––

    l = 84 - 2w

    l = 84 - 2(21)

    l = 42

    length that maximizes area  = 42 ft <–––––


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  • 3 weeks ago

    Michael’s answer is absolutely correct. I am not sure why it would receive two down votes but he was able to answer the question by applying non-calculus methods. He recognized that one of the givens for the area forced the requirement that one side of the shape to be used was not necessary. How you ask? Well in my opinion I suspect he recognized it as a classic maximum/minimum Calculus problem or he looked at the formula for the perimeter which should be 84=2L+2W.  Solving for L we get,

    (84-2W)/2=L.  Or L=42-W.  Next place this value for L in to the equation for the Area of our rectangle to get 

    A=L*W=(42-W)W=A(w).  However, the question set the Area as A(w)=w(84-2w) thus L is set to (84-2w). Using this to determine that the “L” must be only one side comes from, “P“ must be the total length of fence so, 84-2W=L gives 84=L+2W. 

    We have a three sided fenced in Area. If we assume that we want a maximum area then as Michael shows the width will be 1/2 of the length. A=42 x 21 = 882 square feet. Any other combination will only reduce the total area. 

    Applying Calculus to this problem shows the power of Calculus and gives the same answer.  So why the down votes?

    • Michael
      Lv 7
      3 weeks agoReport

      I suspect the down votes are from assuming the classic "max rectangle is a square" solution.
      The give away is A = W(84 - 2W). As you observed, it must be that L = 84 - 2W.
      I used basic algebra rather than the derivative of the area as the vast majority of users in Homework Help are on that level.

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  • 1 month ago

    2L+2W = 84 so L = 42-W

    A = 84W-2W²

    dA/dW = 84-4W = 0 when W = 21 and A" = -4  so that is a maximum

    L = 42 - 21 = 21 so L = W = 21 and Amax = 882

  • 1 month ago

    One way: if y = ax² + bx + c and a < 0, the maximum happens when x = -b / (2a)

    so multiply out to make the equation have the correct form, find a and b and plug in.  That gets the w which gives the maximum; use it to get  the length.

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  • 1 month ago

    This is an interesting change on the traditional problem of fencing in a puppy. Put some numbers in to figure the area and see what happens when you change the numbers.

    Hint: The closer the length and width are to being equal, the larger the area. 

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