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A piece of string 30 inches long has its two ends joined together and is stretched by 3 pegs so as to form a triangle. What is the largest triangular area that can be enclosed by the string?

I took P = a+b+c and Area = 1/2 (b * h) And found h in terms of a and b.Substituting this value to the h in the area equation I differentiated the Area wrt to b and equated to zero.Is this the right way to do it? Can this be soved only using heron's forumla(not A = 1/2(base*height)?

Prnv
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    This question is answered in this post http://math.stackexchange.com/questions/414668/maximize-area-of-a-triangle-with-fixed-perimeter – Miz Jun 05 '15 at 08:51
  • Use differential equations ... I guess the answer is $43.3013$ ... as i guess it is an eq. triangle – NeilRoy Jun 05 '15 at 09:00
  • @AloysiusGodinho I was looking for a solution that uses 1/2 base x height formula.Can we solve it using that equation? – Prnv Jun 05 '15 at 09:10
  • @NeilRoy Yeah the answer is 25 (Sqrt 3).But I can't figure out how to approach the question though.I took P = a+b+c and Area = 1/2 (b*h) And found h in terms of a and b.Substituting this value to the h in the area equation I differentiated the Area wrt to b and equated to zero.Is this the right way to do it? – Prnv Jun 05 '15 at 09:15
  • @NeilRoy: Sure, by symmetry the triangle should be equilateral. But we really need to prove that, rather than merely assuming it. – PM 2Ring Jun 05 '15 at 11:06
  • @Prnv: Heron's formula is the most straight-forward way to attack this problem. Incidentally, how did you go about finding $h$ in terms of $a$ and $b$? It would be good if you gave the details for that in your question. – PM 2Ring Jun 05 '15 at 11:09
  • @PM2Ring http://cl.ly/image/3A2Q2o0z1a2n – Prnv Jun 05 '15 at 11:17
  • @Prnv: Sorry, but that's only valid if $a=c$. Otherwise, the altitude of the triangle will not bisect the base. – PM 2Ring Jun 05 '15 at 11:26
  • @PM2Ring yeah i know but i guess we should look at the link of Aloysius Godinho – NeilRoy Jun 05 '15 at 14:37
  • @PM2Ring Hey if we consider it as a right triangle(a = c),Can we find the solution using the 1/2 b*h? – Prnv Jun 08 '15 at 10:03

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If you are convinced without proof that there is a triangle of maximal area it is easy to see that the maximal triangle has to be equilateral: Keep two pegs $A$, $B$ fixed and use the third peg as a pencil to draw an arc of an ellipse $\gamma_{AB}$ with foci $A$, $B$. The triangle $ABC$ will have maximal area when $C$ is a minor vertex of $\gamma_{AB}$, and will have strictly smaller area for all other positions of $C$. It follows that the triangle of maximal area having a given perimeter is isosceles over all three sides, whence equilateral.

Christian Blatter
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Let $A$ be the area of the square of the required triangle.

Let $s=\dfrac{d}{2}=$half of perimeter of the triangle while $x,y,2 s-x-y$ are the lengths of the sides.

Then

$A=s(s-x)(s-y)(s-2 s+x+y)=s(x-s)(x-y)(x+y-s)$.

Taking partials of A and equating them with 0 we get $x=y$. It follows that the triangle is equilateral and so the area is $\dfrac{1}{4}\dfrac{d}{3}^2\sqrt{3}$, where $d$ is the perimeter.

NeilRoy
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Adelafif
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  • Hey,sorry for the newbie question but what do you mean by taking partials of A here? Also is the only way to solve this question is using the Heron's formula? I tried to solve this using 1/2(b * h) but wasn't able to get to a solution.Can we solve it by using the 1/2(base*height) – Prnv Jun 05 '15 at 10:54
  • @Prnv: Have you learned about partial differentiation yet? – PM 2Ring Jun 05 '15 at 11:04
  • @PM2Ring No, sorry.It is 2 chapter away from this exercise.I am only introduced to find the maxima and minima of functions. – Prnv Jun 05 '15 at 11:06
  • @Prnv: Ok. That makes things a bit trickier, because 2 of the sides can vary independently. However, one of the answers in the question linked by Aloysius Godinho shows how you can deal with that. The standard way to handle maxima & minima with multiple independent variables involves partial differentiation, but that linked answer essentially does an informal version of what you'd do formally with partial differentiation. – PM 2Ring Jun 05 '15 at 11:15
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Look Thompson said that we can solve it by using the hints given on the last line of some page so I'll solve it using that only , using herons formula $A = \sqrt{S(S-a)(S-b)(S-c)}$ , $S$ here is half of perimeter $a,b$ and $c$ are the sides, now u need the maximum value of the part $(S-a)(S-b)(S-c)$. Adding these $(S-a) + (S-b) + (S-c) = 3S - (a+b+c) = 3(15) - (30) = 15$. Now u need to divide 15 in three parts in such a way that u get the maximum value when u multiply the three together, the hint on that page helps us in figuring out that value and its $(n+m+p)/3$ , so $15/3 = 5$. This means that $(S-a) = 5$ , $(S-b) = 5$ and $(S-c)=5$ too. $S=$ half of perimeter = $15$ . sub the values yourself to get $a=b=c=10$ , area of equilateral triangle=$\sqrt{3}/4 \text{(side)}^2$ which gives us the answer

Pizza Thug
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  • Ignore the genius who down voted this. Thompson himself solved it in this way in the chapter "partial differentiation" – Pizza Thug Jul 22 '17 at 12:45