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This is the question

Q) A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is:

The way I tried solve is taking $x$ length of wire to make the square which means $20-x$ is the perimeter of the hexagon.

So, For The sum of Areas $=$ Minimum

$$\frac d{dx} \left[\left(\frac{x}{4}\right)^2 + \frac{3√3}{2} \left(\frac{20-x}{6}\right)^2\right] = 0$$

From this we get $x = -(40√3 +80) $

And if we put $x$ in $\frac{20-x}6$, we get $\frac{50+20\sqrt3}3$

Which isn't even in the options and, I'm unable to detect my mistake here, any help would be much appreciated.

The other is taking $4a + 6b =20$

Which gives the correct value of $b$ which is $\frac{10}{3+2\sqrt3} $

Blank
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  • Welcome to MSE. Here's how to ask a good question, follow these guidelines to get help in this forum. Specifically for this question: – jjagmath May 20 '23 at 11:50
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    "The key parts of your question should not be in an embedded image such as a photo/screenshot/etc., such images are not searchable, they are not accessible to those using screen readers, and they don't show up in question summaries.” – jjagmath May 20 '23 at 11:50
  • Learn to use MathJax to type the math in your questions. Many users in this forum won't even read your question if it's not properly formatted, let alone help you. – jjagmath May 20 '23 at 11:52
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    Choose better titles. "There are 2 same ways to solve this but on give the wrong result which shouldn't be possible." says nothing about the subject of the question. – jjagmath May 20 '23 at 11:53

2 Answers2

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Your approach has a couple of problems.

It looks like your formulas for the area are wrong. The area of a hexagon is $\frac{3\sqrt{3}}{2}a^2$, where $a$ is a single side length. But since $20 - x$ represents the sum of the side lengths of the hexagon, you need to take a sixth of it.

Also, note your functions are defined in terms of the perimeter the square. So you'll be finding the square perimeter length, not the hexagon side length, that minimizes the area of the two shapes.

What I would do is re-define x to be the length of a single side of the hexagon. Then you get $(\frac{20-6x}{4})^2 = (5-\frac{3}{2}x)^2$ to be the function that gives you the area of the square, and $\frac{3\sqrt{3}}{2}x^2$ to be the area of the hexagon.

Now the function $A(x) = (5-\frac{3}{2})^2 + \frac{3\sqrt{3}}{2}x^2$ is the function for the total area of both shapes, in terms of the length of one side of the hexagon. So now you need to find the minimum, so we need to solve $A'(x) = 0$, which, if you work it out, should give you $\frac{10}{3+2\sqrt{3}} \approx 1.547$

Christian Dean
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  • Sorry about the area formula I just typed it out wrong here. And x is defined as the length of wire used to make the square not it's side, therefore I used 20-x as the length of wire used to make the hexagon. – Blank May 22 '23 at 05:24
  • I find your approach better but I wanted to know why mine is wrong. – Blank May 22 '23 at 05:33
  • @Blank Sure. Just for future reference, though, it's best not to edit your question after you've received answers since it can invalidate current answers. Anyway, even with your modified formula, your approach is incorrect because your formula has x representing the perimeter of the square. So when you take the formula's derivative, you've found the value for the perimeter of the square that minimizes the areas of the shapes, but the question was asking for the hexagon side length that minimized the area. – Christian Dean May 22 '23 at 05:49
  • Sorry for the confusion in the original question I thought I wrote it here the way I solved on paper. But could you please give me some examples to why the area is minimized when I take out derivatives of the perimeters. – Blank May 23 '23 at 07:18
  • Hi @Blank, I'm not really sure what you mean to be honest. – Christian Dean May 25 '23 at 23:56
  • Please could you try solving this question by assuming x length to be of the perimeter of the square. I have tried multiple times and failed to annalise my mistake in my approach. – Blank May 26 '23 at 13:44
  • @Blank That approach isn't going to work, though, because the question is asking for you to minimize the area with respect to the hexagon side length, right? So minimizing with respect to another value might happen to give you the right answer, but probably not. So your mistake is your overall approach. – Christian Dean May 26 '23 at 16:23
  • @Blank See how our functions produce two different graphs? https://www.desmos.com/calculator/bdmg5g9y5k The x value that minimizes my function is a different value than the x value that minimizes yours because our functions are minimizing the area with respect to two different variables. – Christian Dean May 26 '23 at 16:26
  • Interesting, so the minimum length of hexagon or square can differ for the same minimum combined area. I mean the minimum combined area would have two possibilities? – Blank May 27 '23 at 12:44
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    @Blank Well in the case we're talking about, the minimum combined areas are the same, but the x-values that minimize them are differentbecause the formulas have $x$ representing different things. That's why your approach doesn't quite work. – Christian Dean May 27 '23 at 14:50
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In your first approach, $$\frac{d}{dx} \left[\left(\frac{x}{4}\right)^2 + \frac{3√3}{2} (20-x)^2\right] = 0$$ should be $$\frac{d}{dx} \left[\left(\frac{x}{4}\right)^2 + \frac{3√3}{2} \left(\frac{20-x}{6}\right)^2\right] = 0.$$

Selrach Dunbar
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    It would have been nicer to complete your solution and to point out where the OP made a mistake. – NoChance May 20 '23 at 04:18
  • Sorry yes you are right I forgot it writes it here . I have solved with the correct f(x) which doesn't give the right answer – Blank May 22 '23 at 04:59