Optimzation problem, only info given are xy intercepts

[weirdobomb]

Homework Statement

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?

Homework Equations

for 2. quotient rule

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).

 

[Mark44]

Homework Statement

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

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.

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?

Homework Equations

for 2. quotient rule

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).

 

[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

 

[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

Am I going in the right direction?

 

[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?

 

[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 y= y_0- (y_0/x_0)x. x_0 is NOT a point on the graph it is <b>number</b>- the x-value of the x-intercept. The &quot;derivative of S&quot;, since it is a line, is the slope of the line, y_0/x_0.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I think fixed point P(a,b) means that, plug in &quot;a&quot; into equation L results with &quot;a.&quot; </div> </div> </blockquote> No, it means that if you put x= a into the equation of the line, you get y= b.