# Find maximum area of a triangle

• MHB
• anemone
In summary, to find the maximum area of a triangle, you can use Heron's formula which takes into account the three side lengths. The third side length is necessary in order to accurately calculate the maximum area. The shape of a triangle does not affect its maximum area, but for a given set of side lengths, the maximum area is achieved when the triangle is equilateral. The maximum area cannot be found using only the angles of the triangle, but there are various methods and algorithms that can be used, such as Heron's formula, trigonometric functions, and calculus. Each method has its own advantages and may be more suitable for different types of triangles.
anemone
Gold Member
MHB
POTW Director
Given two moving points $A(x_1,\,y_1)$ and $B(x_2,\,y_2)$ on parabola curve $y^2=6x$ with $x_1+x_2=4$ and $x_1\ne x_2$ and the perpendicular bisector of segment $AB$ intersects $x$-axis at point $C$. Find the maximum area of $\triangle ABC$.

This seems to be one of those problems where the calculation is easier if you make it more general. I shall take the parabola to be $y^2 = 4ax$ (so eventually $a$ will become $\frac32$), and the condition on the points to be $x_1+x_2 = k$ (with $k=4$ eventually).

Let $A$ and $B$ be the points $(as^2, 2as)$, $(at^2,2at)$, so that $as^2 + at^2 = k$. The line $AB$ has slope $\dfrac2{s+t}$ and midpoint $\bigl(\frac12k,a(s+t)\bigr)$. Its perpendicular bisector therefore has equation $y - a(s+t) = -\tfrac12(s+t)(x - \tfrac12k)$, and it meets the $x$-axis at the point $(2a+\tfrac12k,0)$. The area $\Delta$ of triangle $ABC$ is the absolute value of $$\frac12\begin{vmatrix}as^2&2as&1\\ at^2&2at&1\\ 2a+\frac12k&0&1 \end{vmatrix} = (2a+\tfrac12k)a(s-t) + a^2st(s-t).$$ Now let $u = s-t$ and $v = s+t$. Then $u^2+v^2 = 2(s^2+t^2) = \frac{2k}a$ and $v^2-u^2 = 4st$, so that $st = \frac14(v^2-u^2) = \frac12\bigl(\frac ka - u^2\bigr)$. Therefore $$\Delta = \left| (2a+\tfrac12k)a(s-t) + a^2st(s-t) \right| = \left| (2a+\tfrac12k)au + a^2\tfrac12\bigl(\tfrac ka - u^2\bigr)u\right| = \left| (2a+k)au - \tfrac12a^2u^3 \right|.$$ Now plug in the values $a=\frac32$ and $k=4$ to get $\Delta = \left| \frac{21}2u - \frac98u^3\right|$. Also, $u^2 \leqslant u^2+v^2 = \frac{2k}a = \frac{16}3$, so that $u$ must lie in the interval $\left[-\frac4{\sqrt3},\frac4{\sqrt3}\right]$. The maximum value of $\left| \frac{21}2u - \frac98u^3\right|$ in that interval occurs at the turning point when $\frac{21}2 - \frac{27}8u^2 = 0$ and so $u = \frac{2\sqrt7}3$. Therefore the maximum value of $\Delta$ is $\Delta_{\max} = \frac{2\sqrt7}3\left(\frac{21}2 - \frac98\frac{28}9\right) = \frac{14}3\sqrt7 \approx 12.347$.

Having found $u$, it is easy to work out corresponding values of $s$ and $t$, which can be taken as $\frac13(\sqrt7\pm\sqrt5)$. The points $A$, $B$ and $C$ are then $$A = \left( 2 - \tfrac13\sqrt{35}, \sqrt7 - \sqrt5\right) \approx (0.08,0.41),$$ $$B = \left( 2 + \tfrac13\sqrt{35}, \sqrt7 + \sqrt5\right) \approx (3.97,4.88),$$ $$C = (5,0).$$
[TIKZ]\draw [help lines, ->] (-1,0) -- (6.5,0) ;
\draw [help lines, ->] (0,-3) -- (0,6.5) ;
\draw [domain=-1:2, samples=100] plot ({1.5*\x*\x}, {3*\x });
\coordinate [label=left:$A$] (A) at (0.03,0.41) ;
\coordinate [label=above left:$B$] (B) at (3.97,4.88) ;
\coordinate [label=below:$C$] (C) at (5,0) ;
\draw [thick] (A) -- (B) -- (C) -- cycle ;
\draw (2,2.65) -- (C) ;
\foreach \x in {2,4,6} { \draw (\x,-0.1) node [ below ] {$\x$} -- (\x,0.1) ;
\draw (-0.1,\x) node [ left ] {$\x$} -- (0.1,\x) ; } ;[/TIKZ]

## 1. How do you find the maximum area of a triangle?

To find the maximum area of a triangle, you can use the formula A = 1/2 * base * height. The base and height are the two sides of the triangle that form a right angle. You can also use the Heron's formula, which takes into account all three sides of the triangle.

## 2. What is the importance of finding the maximum area of a triangle?

Finding the maximum area of a triangle is important in various fields such as engineering, architecture, and mathematics. It helps in determining the most efficient use of space and resources, and can also be used to solve optimization problems.

## 3. What are some methods for finding the maximum area of a triangle?

Aside from using the basic formula and Heron's formula, there are other methods for finding the maximum area of a triangle. These include using trigonometric functions, calculus, and geometric constructions.

## 4. Can the maximum area of a triangle be greater than its perimeter?

Yes, it is possible for the maximum area of a triangle to be greater than its perimeter. This can occur when the triangle is obtuse, meaning one of its angles is greater than 90 degrees. In this case, the base and height of the triangle will be longer than any of its sides.

## 5. How can I apply the concept of finding the maximum area of a triangle in real-life situations?

The concept of finding the maximum area of a triangle can be applied in various real-life situations, such as determining the most efficient shape for a piece of land, finding the maximum capacity of a container, or optimizing the design of a bridge or building. It can also be used in calculating the maximum profit for a business or the maximum yield for a crop field.

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