# 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

Relevance

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

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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 <––––– • the question as asked does not specify the maximum area as a condition of the solution..

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• 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?

• 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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• 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

• Because you have LxW wrong.

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• 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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