Difficult Optimization

[Kushballo7]

Alright so I am completely lost on this problem.

The line y = mx+b intersects the parabola y = x^2 at points A in the second quadrant and B in the first quadrant. Find the point P on the arc AOB where O is the origin, that maximizes the area of triangle PAB.

 

[pka]

I think this is a very subtle problem.
Consider the length of the line segment AB call it j.
Then j is the base of the triangle.
Lets say that P has coordinates (p,p<SUP>2</SUP>), then the distance from P to the line AB is the altitude of the triangle.
The line AB has equation mx−y+b=0, so the distance is $\displaystyle \frac{{\left| {mp - p^2 + b} \right|}}{{\sqrt {m^2 + 1} }}$.
Therefore, the area of the triangle is a function of p,
$\displaystyle A(p) = \frac{{\left| {mp - p^2 + b} \right|}}{{2\sqrt {m^2 + 1} }}j$ ( recall j is the length of the base).

Note that $\displaystyle A'(p) = \frac{{m - 2p}}{{\left( {2\sqrt {m^2 + 1} } \right)\left| {mp - p^2 + b} \right|}}j$.

Thus the maximum occurs if $\displaystyle p = \frac{m}{2}$.

 

[Kushballo7]

I think this is a very subtle problem.
Consider the length of the line segment AB call it j.
Then j is the base of the triangle.
Lets say that P has coordinates (p,p<SUP>2</SUP>), then the distance from P to the line AB is the altitude of the triangle.
The line AB has equation mx−y+b=0, so the distance is $\displaystyle \frac{{\left| {mp - p^2 + b} \right|}}{{\sqrt {m^2 + 1} }}$.
Therefore, the area of the triangle is a function of p,
$\displaystyle A(p) = \frac{{\left| {mp - p^2 + b} \right|}}{{2\sqrt {m^2 + 1} }}j$ ( recall j is the length of the base).

Note that $\displaystyle A'(p) = \frac{{m - 2p}}{{\left( {2\sqrt {m^2 + 1} } \right)\left| {mp - p^2 + b} \right|}}j$.

Thus the maximum occurs if $\displaystyle p = \frac{m}{2}$.

I do not under stand how you got the distance part.

 

[pka]

“I do not under stand how you got the distance part.”
I assume that you mean that you do not understand the formula for the distance of a point to a line. You should look up this up in your text.

I can give you a general answer.
Given a point (p,q) and a line in standard form Ax+By+C=0 then the distance from that point to the line is give by:

$\displaystyle \frac{{\left| {Ap + Bq + C} \right|}}{{\sqrt {A^2 + B^2 } }}$.

 

[Kushballo7]

“I do not under stand how you got the distance part.”
I assume that you mean that you do not understand the formula for the distance of a point to a line. You should look up this up in your text.

I can give you a general answer.
Given a point (p,q) and a line in standard form Ax+By+C=0 then the distance from that point to the line is give by:

$\displaystyle \frac{{\left| {Ap + Bq + C} \right|}}{{\sqrt {A^2 + B^2 } }}$.

Yeah, after I posted that I found it online. Thanks for your help!!

 

[Gene]

I have a slightly different approach. AB is the base of the triangle so the max area is when the height is max. That will be when P is on a tangent parallel to the line mx+b. The tangent is 2x. So
2x=m and P is at (m/2,(m/2)²)
Not so hard unless someone can shoot it down :twisted:

 

[pka]

Gene, That is good!

 

[galactus]

You fellas approaches are very clever. I'm going to give it a shot. If it ain't right it'll be wrong, I suppose.

The vertical line of length C divides the triangle into 2 smaller triangles which share the same base length C. The heights of the two triangles are x-a and b-x.

So, the area of the big triangle is:

$\displaystyle Area=(x-a)C/2+(b-x)C/2=(b-a)C/2$

As you both pointed out, the area is max when C is max.

The slope of the line is$\displaystyle \frac{(b^{2}-a^{2})}{(b-a)}=(a+b)$

So it's equation, in point-slope form, is:

$\displaystyle y-a^{2}=(a+b)(x-a),$

$\displaystyle y=(a+b)\cdot(x)-ab$.

$\displaystyle C=(a+b)\cdot(x)-ab-x^{2},$ if$\displaystyle a\leq x \leq b$

$\displaystyle \frac{dC}{dx}=a+b-2x$

Setting to 0 and solving for x we get $\displaystyle x=\frac{a+b}{2}$

So P should be at: $\displaystyle x=\frac{(a+b)}{2}$ and $\displaystyle y=\frac{(a+b)^{2}}{4}$

Sub x into C and we have: $\displaystyle C=\frac{(b-a)^{2}}{4}$

the area of the triangle is greatest when C=$\displaystyle \frac{(b-a)^{2}}{2}$

the area is $\displaystyle \frac{(b-a)^{3}}{8}$

 

[Gene]

It's nice to be verified.
So P should be at: x = (a+b)/2 = m/2
----------------
Gene