A line and area of a triangle

Here's a statement Area of triangle formed by the line passing through the point (α,β)\left( \alpha ,\beta \right), with axes is maximum when point (α,β)\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)A(0,n) , B(α,β)B(\alpha,\beta), C(h,0)C(h,0) and AB:BC::k:1AB:BC::k:1

We know, α=khk+1&β=nk+1\alpha =\frac { kh }{ k+1 } \quad \& \quad \beta =\frac { n }{ k+1 }

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

Area of ΔAOC\Delta AOC, A=12×α(k+1)k×β(k+1)A=\frac { 1 }{ 2 } \times \frac { \alpha \left( k+1 \right) }{ k } \times \beta \left( k+1 \right)

Differentiating w.r.t k,

dAdk=12αβ[2k(k+1)(k+1)2k2]\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 dAdk\frac { dA }{ dk } to zero k=±1\Rightarrow k=\pm 1

Maxima occurs at k=1k=1

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

k=1k=1\Rightarrow (α,β)(\alpha,\beta) is the midpoint of AA and CC.

Hence, Statement is CORRECT.

Note by Anandhu Raj
5 years, 3 months ago

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1 vote

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

Note that xa+yb=1 \dfrac{x}{a} + \dfrac{y}{b} = 1 is the equation of the line in intercept form. a a and b b are the x x and y y intercepts respectively.
We need to maximize =ab2 \triangle = \dfrac{ab}{2} .
As α,β \alpha, \beta lies on the line, αa+βb=12αβ2 \dfrac{\alpha}{a} + \dfrac{\beta}{b} = 1 \geq 2 \sqrt{\dfrac{\alpha \beta}{2 \triangle}}
2αβ \Rightarrow \triangle \geq 2 \alpha \beta with equality when αa=βb=12 \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 - 5 years, 3 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 dAdk\frac { dA }{ dk } and green curve represent d2Adk2\frac { { d }^{ 2 }A }{ d{ k }^{ 2 } }

Anandhu Raj - 5 years, 3 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 xa+yb=1\frac { x }{ a } +\frac { y }{ b } =1 and xb+ya=1\frac { x }{ b } +\frac { y }{ a } =1, meets the coordinate axes in AA andBB. Then find the locus of the midpoint of ABAB.

Anandhu Raj - 5 years, 3 months ago

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