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A line and area of a triangle

Here's a statement Area of triangle formed by the line passing through the point \(\left( \alpha ,\beta \right)\), with axes is maximum when point \(\left( \alpha ,\beta \right)\) is the midpoint of the line between the axes.

The question was to answer whether the statement was correct or not and the answer given is that the statement is WRONG.

But why?

Here's what I did,

Let \(A(0,n)\) , \(B(\alpha,\beta)\), \(C(h,0)\) and \(AB:BC::k:1\)

We know, \[\alpha =\frac { kh }{ k+1 } \quad \& \quad \beta =\frac { n }{ k+1 } \]

\[\Rightarrow h=\frac { \alpha \left( k+1 \right) }{ k } \quad \& \quad n=\beta \left( k+1 \right) \]

Area of \(\Delta AOC\), \[A=\frac { 1 }{ 2 } \times \frac { \alpha \left( k+1 \right) }{ k } \times \beta \left( k+1 \right) \]

Differentiating w.r.t k,

\[\Rightarrow \frac { dA }{ dk } =\frac { 1 }{ 2 } \alpha \beta \left[ \frac { 2k\left( k+1 \right) -{ \left( k+1 \right) }^{ 2 } }{ { k }^{ 2 } } \right] \]

Equating \(\frac { dA }{ dk } \) to zero \(\Rightarrow k=\pm 1\)

Maxima occurs at \(k=1\)

\(\therefore \) Area is maximum when \(k=1\).

\(k=1\Rightarrow \) \((\alpha,\beta)\) is the midpoint of \(A\) and \(C\).

Hence, Statement is CORRECT.

Note by Anandhu Raj
1 year, 8 months ago

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An application of AM-GM will tell you why you've gotten the wrong answer.  

Note that \( \dfrac{x}{a} + \dfrac{y}{b} = 1 \) is the equation of the line in intercept form. \( a \) and \( b \) are the \( x \) and \( y \) intercepts respectively.
We need to maximize \( \triangle = \dfrac{ab}{2} \).
As \( \alpha, \beta \) lies on the line, \( \dfrac{\alpha}{a} + \dfrac{\beta}{b} = 1 \geq 2 \sqrt{\dfrac{\alpha \beta}{2 \triangle}} \)
\( \Rightarrow \triangle \geq 2 \alpha \beta\) with equality when \( \dfrac{\alpha}{a} = \dfrac{\beta}{b} = \dfrac{1}{2} \)
 

So you have the minimum area, not the maximum.
Be careful when using calculus to find extrema - check minima and maxima always.

Ameya Daigavane - 1 year, 8 months ago

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@Ameya Daigavane Sorry disturb you again, sir. But could you please post a solution to below question?

A variable line, drawn through the point of intersection of the straight lines \(\frac { x }{ a } +\frac { y }{ b } =1\) and \(\frac { x }{ b } +\frac { y }{ a } =1\), meets the coordinate axes in \(A\) and\(B\). Then find the locus of the midpoint of \(AB\).

Anandhu Raj - 1 year, 8 months ago

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Comment deleted Mar 24, 2016

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Comment deleted Mar 24, 2016

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@Anandhu Raj I am extremely sorry. This is what happens when you are in a hurry.
The point of intersection is \( (\frac{ab}{a + b}, \frac{ab}{a + b}) \) ( This was the mistake.)
If \( A \) and \( B \) are the \( x \) and \( y \) intercepts respectively for the new line, our point lies on the line, so, \( \dfrac{\frac{ab}{a + b}}{A} + \dfrac{\frac{ab}{a + b}}{B} = 1 \)

But, \( (x, y) = (\frac{A}{2}, \frac{B}{2}) \) is the midpoint.

So, \( \dfrac{\frac{ab}{a + b}}{2x} + \dfrac{\frac{ab}{a + b}}{2y} = \dfrac{ab}{2(a + b)} \cdot \left(\dfrac{1}{x} + \dfrac{1}{y}\right) = \dfrac{ab}{2(a + b)} \cdot \left(\dfrac{y + x}{xy}\right) =1 \)

which is the answer. Sorry again. I've deleted the other 'solution' to not mislead people.

Ameya Daigavane - 1 year, 8 months ago

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@Ameya Daigavane Thank you, Sir :)

Anandhu Raj - 1 year, 8 months ago

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Oh! yes sir, I made a mistake while checking maxima and minima! And your approach is superb!

The red curve represent \(\frac { dA }{ dk } \) and green curve represent \(\frac { { d }^{ 2 }A }{ d{ k }^{ 2 } } \)

Anandhu Raj - 1 year, 8 months ago

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