# Homework Help: Optimzation problem, only info given are xy intercepts

1. Nov 19, 2012

### weirdobomb

1. The problem statement, all variables and given/known data
Suppose P(a,b) is a fixed point in Quadrant 1 of an xy-plane and line L is descending in the plane such that P is on line L. Let Q=(xknot,0) and R=(0,yknot) be the x and y intercepts for line L and let

S= 1/(xknot + yknot)

1. Express S as a function of xknot
2. Find any extreme values for S

I know how to do number 2 but number 1 stumps me, how do I get yknot to become xknot?

2. Relevant equations

for 2. quotient rule

3. The attempt at a solution
so far I have got L=-(y/x) + y using y=mx+b (descending line means it's linear right?)

in another attempt I find the equation of the line using point RP and then PQ then setting them equal to each other to solve for yknot. I use that to plug into S. S would be expressed in terms of xknot and the a and b are constants. If I find the critical numbers of S with this, it comes out very ugly and does not seem like the answer.

HOW DO I APPROACH THIS PROBLEM (it is supposedly an optimization problem).

2. Nov 19, 2012

### Staff: Mentor

It's pronounced like "knot" but the word is "nought", which is synonomous with zero.

What does it mean that (a, b) is a fixed point? That's key to this problem.

3. Nov 19, 2012

### weirdobomb

I think fixed point P(a,b) means that, plug in "a" into equation L results with "a."
I get

S = (x0-a)/x0^2

Last edited: Nov 20, 2012
4. Nov 19, 2012

### weirdobomb

Am I going in the right direction?

Last edited: Nov 20, 2012
5. Nov 20, 2012

### weirdobomb

Quick question:

since x0 and a are points on the graph, is it say to say that the first derivative of S is 0 since they are constants?

6. Nov 20, 2012

### HallsofIvy

If $P= (x_0, 0)$ and $Q= (0, y_0)$ then the equation of the line is $x/x_0+ y/y_0= 1[/tex] or [itex]y= y_0- (y_0/x_0)x$. $x_0$ is NOT a point on the graph it is number- the x-value of the x-intercept. The "derivative of S", since it is a line, is the slope of the line, $y_0/x_0$.

No, it means that if you put x= a into the equation of the line, you get y= b.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook