# Minimization problem

• Sep 21st 2009, 10:29 AM
pantera
Minimization problem
Hello,
How do I set up the equation that will find the answer for:
Find the x-coordinates of the point P on the curve y = 1/xsquared (x>0)
where the segment of the tangent line at P that is cut off by the coordinate axes has its shortest length.

Thanks
• Sep 21st 2009, 12:42 PM
tom@ballooncalculus
Sketch the function and an example tangent line in the first quadrant.

http://www.ballooncalculus.org/asy/graphs/graph1.png

Label the right-triangle at the corners x0 and y0...

... and at points (x, 0) and (0, 1/x^2) on the axes, that give the x and y values of the point where the tangent touches the curve.

http://www.ballooncalculus.org/asy/graphs/graph2.png

You have 3 similar triangles. So express the equal ratios (that of the big triangle being minus the slope of the tangent) and find x0 and y0 in terms of x.

So get H the hypotenuse in terms of x.

Then see where this function reaches a minimum...
• Sep 22nd 2009, 10:09 AM
pantera
Thanks, but I still didn't get it.
• Sep 22nd 2009, 02:59 PM
mr fantastic
Quote:

Originally Posted by pantera
Thanks, but I still didn't get it.

An alternative approach:

Let P have coordinates $\displaystyle \left(a, \, \frac{1}{a^2} \right)$.

Then the equation of the tangent at P is $\displaystyle y - \frac{1}{a^2} = - \frac{2}{a^3} (x - a)$.

Get the x-intercept (point A). Get the y-intercept (point B). Hence get (in terms of a) the length of the line segment joining A and B using the usual distance formula.

Use calculus in the usual way to minimise the length. (In fact, you will find it easiest to minimise the square of the length. Think about why it's OK to do that). Hence get the value of a.
• Sep 22nd 2009, 03:03 PM
pantera
Thank you.