Straight Lines

  • Thread starter RandomGuy1
  • Start date
  • #1
19
0
Homework Statement

A straight line L with negative slope passes through the point (8,2) and cuts the positive coordinate axes at points P and Q. As L varies, what is the absolute minimum value of OP + OQ?

The attempt at a solution

Let x and y be two points on the line.
Using the slope point formula, we have (y-2) = m(x-8).
Writing this in intercept form, one gets x/(8 - 2/m) + y/(2 - 8m) = 1.
This implies coordinates of P and Q are (8 - 2/m, 0) and (0, 2 - 8m) respectively.

∴ OP + OQ = 10 + (-2/m - 8/m)

I can't figure out how to proceed from there. How do you find out the minimum value for the last term?
 

Answers and Replies

  • #2
hilbert2
Science Advisor
Insights Author
Gold Member
1,517
535
This implies coordinates of P and Q are (8 - 2/m, 0) and (0, 2 - 8m) respectively.

∴ OP + OQ = 10 + (-2/m - 8/m)

Actually [itex]OP + OQ = 10 - 2/m -8m[/itex] . You also need the condition [itex]m<0[/itex] bacause the slope of the line is required to be negative.

If you've learned derivatives at school, you just have to find the extremum values of [itex]OP + OQ[/itex] by solving [itex]\frac{d}{dm}\left(10-2/m-8m\right)=0[/itex] and choose the negative solution to find the correct value of m.
 
  • #3
19
0
Oh, sorry, I meant -8m. Could you please explain the last step? Why do we equate the derivative to zero?
 
  • #4
hilbert2
Science Advisor
Insights Author
Gold Member
1,517
535
Oh, sorry, I meant -8m. Could you please explain the last step? I know the derivative of -8m becomes 8. Wouldn't d/dm (-2/m) simply become -2/m2

Using the rule [itex]\frac{d}{dx}x^{n}=nx^{n-1}[/itex] with n=-1, we get [itex]\frac{d}{dm}(-2/m) =-2\frac{d}{dm}m^{-1}=-2\times(-1)\times m^{-2}=2/m^{2}[/itex]. You just had the wrong sign.

Why do we equate the derivative to zero?

If a differentiable function f(x) has an extremal (smallest or largest) value at some point a, we must have f'(a)=0.
 
  • #5
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
Homework Statement

A straight line L with negative slope passes through the point (8,2) and cuts the positive coordinate axes at points P and Q. As L varies, what is the absolute minimum value of OP + OQ?

The attempt at a solution

Let x and y be two points on the line.
Using the slope point formula, we have (y-2) = m(x-8).
Writing this in intercept form, one gets x/(8 - 2/m) + y/(2 - 8m) = 1.
This implies coordinates of P and Q are (8 - 2/m, 0) and (0, 2 - 8m) respectively.

∴ OP + OQ = 10 + (-2/m - 8/m)

I can't figure out how to proceed from there. How do you find out the minimum value for the last term?

If we let the slope be ##-m## we have ##y-2 = -m(x-8)##, with ##m > 0##. We have
[tex] \text{OP} + \text{OQ} = 10 + 8m + \frac{2}{m} \equiv S(m).[/tex] We want to minimize ##S(m)##, so we need to minimize ##8m + 2/m##. We can use calculus, or we can use a trick involving the arithmetic-geometric mean inequality: for ##a, b > 0## we have
[tex] \frac{1}{2}(a+b) \geq \sqrt{a b} \Longrightarrow a+b \geq 2 \sqrt{ab},[/tex] with equality holding if and only if ##a = b##. So, applying this we have
[tex] 8m + \frac{2}{m} \geq 2 \sqrt{8m \frac{2}{m}} = 2 \sqrt{16} = 8,[/tex] with equality holding only when ##8m = 2/m,## or ##m^2 = 1/4##, or ##m = \pm 1/2 \longrightarrow m = 1/2##.
 

Related Threads on Straight Lines

  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
9
Views
4K
  • Last Post
Replies
5
Views
920
  • Last Post
Replies
5
Views
3K
Replies
6
Views
4K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
1
Views
2K
Top