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The problem is as follows:

Alice and Willy had bought a new house. The figure from below is a sketch of the house. The couple wants the home to be surrounded completely by a wooden fence, which must occupy the maximum area as posible in front of the house. This is is indicated by a lighter shade in the drawing where $ABCD$ is a rectangle. The couple however, only got the materials including the wood, nails and paint from the hardware store to complete $98$ meters of the fence. Assuming the maximum area the couple wants to enclose starts in front of the house as indicated in the arrow up to the entrance of the house, How many meters in a straight line from the house will the entrance to the fence be located?

Sketch of the problem

The alternatives in my book are as follows:

$\begin{array}{ll} 1.&18\,m\\ 2.&14\,m\\ 3.&16\,m\\ 4.&25\,m\\ \end{array}$

I'm assuming that to solve this problem it is required the use of derivatives, since it will be generated a quadratic equation. But I don't know how to establish such equation. Can someone help me with this part?.

Since what they require is to get the maximum area, I believe it will be given by:

$a(b+c+2)=A$

Assuming the small segments b and c are between the entrance door in the fence.

Since $a+b+c=98$

Then this means:

$a(98-a+2)=A(a)$

Then this would mean:

$A(a)=a(100-a)=100a-a^2$

Since the function will be as follows:

$A(a)=100a-a^2$

Thus the way to attain the maximum will be the derivate equal to zero.

$A'(a)=100-2a=0$

$a=50$

Therefore this would be the maximum.

But this answer does not appear in the alternatives. Which part did I made a mistake?. Can someone help me here?. Please an answer must include a drawing since for me it is difficult to spot where to establish the right equation in order to maximize it.

Chris Steinbeck Bell
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  • The question as posed is nonsense: they seem to want you to assume that $BC$ is variable, but the picture says otherwise! As for your answer, you have evaluated $98+2$ as $96$... – TonyK Nov 01 '20 at 03:05
  • @TonyK see my answer. Yes, the picture is extremely misleading. The garbled intent is that the width of the fence is variable. – user2661923 Nov 01 '20 at 03:08
  • @user2661923 I'm sorry if the drawing seems misleading. Or maybe it is contradicting the statement in the question. What it was intended to be asked is that they want to have the maximum area as possible assuming to cover from the front door of the house. The other thing about $BC$ it is just as taking it as a reference point but not as a changing parameter. This part would not be covered by the fence. – Chris Steinbeck Bell Nov 01 '20 at 04:34
  • @TonyK Sorry. I didn't noticed that. But it seems that if correcting this would had meant: $A(a)=100a-a^2$, hence; $A'(a)=100-2a=0$, but I don't know if this $50,m$. But this also doesn't help much to the solution. – Chris Steinbeck Bell Nov 01 '20 at 04:43
  • @TonyK Hi. I have edited the question and a misleading line. Maybe is it clearer now?. As it stands can this problem be solved?. – Chris Steinbeck Bell Nov 01 '20 at 04:47
  • I have added an Addendum to my answer to cover your questions. If more questions, either add a comment directly after my answer, or address the comment to @user2661923. – user2661923 Nov 01 '20 at 05:09
  • I don't see how your edit has changed anything. The picture is still an incompetent misrepresentation of the question. – TonyK Nov 02 '20 at 14:11

1 Answers1

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Your formula for the area is wrong.

Suppose that the side lengths AB and CD are each $x$ units.

Then, the width of the rectangle will be

$$98 - (2 \times x) + 2 = 100 - 2x.$$

This means that the area will be

$$f(x) = (100 - 2x) \times x = 100x - 2x^2.$$

Taking the derivative,

$$f'(x) = 100 - 4x.$$

This means that $f(x)$ will have a maximum when

$$100 - 4x = 0.$$

Addendum
Per OP's request:
Reactions to his recent questions.

First, see my comment that immediately follows your query.

I adopted the interpretation that the width of the fence is variable, only because that was the only way that the problem could be attacked.

Let's take it one step at a time.

There are two possibilities:

  • either the width of the fence is some fixed distance a.

  • or the width of the fence is variable.

If the width of the fence is fixed, at the value $a$, then the problem makes no sense! Naturally, you would then use all of the fence, so the length of each of the sides would be

$$\left(\frac{1}{2}\right) \times \left[98 - (a-2)\right] ~=~ \left[50 - \left(\frac{a}{2}\right)\right].$$

Since the assumption of a fixed-width fence leads to the conclusion that the dimensions of the fence are fixed, the assumption is untenable.

Assuming that the dimensions of the width of the fence equals the variable $a$, and that the dimensions of each side of the fence is $x$, then you have that

$$(a-2) + 2x = 98 \implies a = (100 - 2x).$$

Therefore, the problem is reduced to only dealing with one variable, $x$.

You have raised a separate question - why couldn't the problem be attacked using two variables, $a$ and $x$. Although this is feasible, as long as you remain aware that $a$ must equal $(100 - 2x)$ it adds an unnessary complication.

The main reason that you had trouble relates to the first sentence that I posted in my original answer: your formula for the area is wrong.

You had

$$a(b + c + 2) = A. \tag1$$

I soon as I saw that, I stopped reading your analysis, because that formula for the area makes no sense. If AB and CD have lengths $b$ and $c$ respectively, where $b = c$, then the area is

$$A = (a \times b).$$

If (for some bizarre reason), $b \neq c$, then you have a trapezoid and the area is

$$A = \left(a \times \frac{b+c}{2}\right).$$

Therefore, there was no way that your formula in line (1) above, was correct.

You raised another point:

The other thing about BC it is just as taking it as a reference point but not as a changing parameter.

I always assumed that points B and C were fixed. The variable $x$ was intended to represent the length of line BA, which I presumed to be equal to the length of line CD.

Another question

As it stands can this problem be solved?

My original answer stands; I have already solved the problem. The key points in my solution are

  • The width of the fence is variable, rather than being the fixed length of line BC.

  • The problem can be reduced to a single variable problem (which is clearly best) by understanding that the width $a$ must equal $(100 - 2x).$

user2661923
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  • Why are you assuming that the width of the rectangle to be $98-(2x)+2$?. Why? can you help me with this part?. I find it confusing. Why did using two different variables for each side doesn't seem to work? – Chris Steinbeck Bell Nov 01 '20 at 04:29
  • I'm sorry for the inconvenience. I had to translate this question from the original source which is in spanish. The information regarding $BC$ may be omitted, as indicated in the comments above the couple intended to fill the greatest area possible in front of their home using only $98,m$ of fence. But I'm still confused on why $x$ is the same for both sides?. This alludes a square, which isn't the case. Maybe you want to add a clarification attending this question I have because that part I find it confusing. – Chris Steinbeck Bell Nov 01 '20 at 04:42
  • I have edited my question and removed any misleading clue in the question. I do hope it is clearer now. I will really appreciate you could see my changes and answer my pending comments. – Chris Steinbeck Bell Nov 01 '20 at 04:48
  • Howdy!. I've been reading the addendum you put but i'm stuck. You say that if we assume the width of the fence be fixed yo some value calling it $a$ this would lead us to $50-\frac{a}{2}$ will be the length of the sides. This part is confusing, what are you intending to refer with sides?. Do you refer to the sides contiguous to the entrance gate I mean those on $AD$? and are you establishing that they are equal?, so this is the reason why is it taken its half?. Can you make a clarification for this?. – Chris Steinbeck Bell Nov 02 '20 at 04:22
  • I'm getting the idea that $BC=a$ and $AB=CD=x$ is this correct? By following your explanation I could understand why did you got the dimentions for the width and the lenght. It made sense that the portion at both sides of the entrance gate are the same hence adding these or just using $a-2$ would give the all length of the fence. This equation can allow having the area equation in a single variable. I'm still stuck with what I mentioned lines above, can you please attend that question?. – Chris Steinbeck Bell Nov 02 '20 at 04:22
  • Yes, the idea that I was expressing was that if the width was fixed at $a = $ BC, then the side dimension $x =$ AB = CD would also be fixed by $2x + (a-2) = 98.$ This means that if the width $a$ is fixed, the query makes no sense, since then, both the width and the depth (i.e. both $a$ and $x$) of the enclosed rectangular area would be fixed. This is clearly contrary to the intent of the question which is asking you to determine which dimensions for width and depth may be used to maximize the enclosed area. – user2661923 Nov 02 '20 at 04:29
  • @ChrisSteinbeckBell see comment above – user2661923 Nov 02 '20 at 04:30