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)?
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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.
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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
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.