Minimizing the area in a parabola

[r3dxP]

Let f(x) = 6-x^2 for 0<w<sqrroot(6); let A(w) be the area of the triangle formed by the coordinate axes and the line tangent to the graph of f.
see figure : http://img472.imageshack.us/img472/50/a5nt.jpg

 

[HallsofIvy]

Okay, I looked at your picture and read your post. There is no question there! I can guess from the title that the problem is to find the point on the parabola such that the triangle constructed has minimum possible are- though that is NOT "minimizing the area in a parabola"!

Have you done anything at all on this? In particular, taking the "x" coordinate of the point on the parabola as fixed (call it x₀, say) can you write the equation of the tangent line there?
Once you have that, you should be able to write the equation for the area of the triangle as a function of x₀.

 

[r3dxP]

Okay, I looked at your picture and read your post. There is no question there! I can guess from the title that the problem is to find the point on the parabola such that the triangle constructed has minimum possible are- though that is NOT "minimizing the area in a parabola"!
Have you done anything at all on this? In particular, taking the "x" coordinate of the point on the parabola as fixed (call it x₀, say) can you write the equation of the tangent line there?
Once you have that, you should be able to write the equation for the area of the triangle as a function of x₀.

yea i did try it like 20x but couldn't figure it out..
i derived to get the tangent line, so slope = -2x at point x..
so i find the equation for the line.. thus finding b as 6+x^2
i got this by plugging (x,6-x^2) into the equation y = -2x(x) + b and solved for b.
then i get the final equation of y = 6-x^2! i end up where i started from! what is going on?!

 

[TD]

The equation for a tangent line to a curve f(x) at a point (a,b) is given by y-b = f'(a)(x-a). Note that with f'(a), I mean the derivative of f(x) evaluated in the point (a,b) so with x = a.
So instead of (a,b), you know have the general point (p,6-p²). You already had the derivative, so can you know get to the tangent line for a general point on the parabola? It's just filling in.

After that, how would you find the area of the triangle? You're "lucky" since it's a right triangle ;)

 

[r3dxP]

if i just find the tangent line equation, i end up with y=6-x^2 as my line of my tangent line. Is that suppose to be right?
and how am i suppose to find the minimum area under the triangle and prove that its the smallest possible area of the triangle on the parabola function that's given?

 

[TD]

if i just find the tangent line equation, i end up with y=6-x^2 as my line of my tangent line. Is that suppose to be right?
and how am i suppose to find the minimum area under the triangle and prove that its the smallest possible area of the triangle on the parabola function that's given?

In the equation I gave you, don't fill in (x,6-x²) since x is also your unknown in the equation, your variable! Instead, use (e.g.) (p,6-p²). Just apply the formula I gave. Then, for every p, you get a point (p,6-p²) on the parabola with the tangent line you just set up, p being substituted with the value in question.

 

[r3dxP]

so.. y-(6-p^2) = f'(p) * (x - p)
y - 6 + p^2 = -2p (x-p)
y = p^2 -2px + 6 ? what does this mean? the equation of the tangent line?

 

[TD]

Correct, for every p you fill in, you'll now get the tangent line to the parabola at the point x = p, y = 6-p^2.

 

[r3dxP]

ok, so.. what i did was.. set 0=-2px + p^2 + 6, then solved for x. x= (p^2+6)/2p;
where as the y-intercept is (0, p^2 +6)
so to get the area.. A(p) = 1/2 * (p^2+6)/2p * (p^2+6)
A(p) = (p^4 + 12p^2 + 36) / 4p
I derive this area to find the min/max..
am i on the right track?

 

[Galileo]

Yes, that looks good.

One advice though. It's generally easier (in my opinion) to not expand the brackets. So just keep A(p) look like:
A(p)=\frac{(p^2+6)^2}{4p}
Taking the derivative will be easier in this form

 

[r3dxP]

thanks a lot.. i got A'(w) = 0 where w=-6 and 2.
thus A(2) = 12.5; which is the min point I am guessing.
i can check the value of A''(2) to see if its <0 or >0 to det. if its min or max but I am too lazy. i think i did this right right now. once again, thank you all for the help.