Finding Absolute Min Value of OP + OQ with Slope m

[RandomGuy1]

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?

 

[hilbert2]

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 OP + OQ = 10 - 2/m -8m . You also need the condition m<0 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 OP + OQ by solving \frac{d}{dm}\left(10-2/m-8m\right)=0 and choose the negative solution to find the correct value of m.

 

[RandomGuy1]

Oh, sorry, I meant -8m. Could you please explain the last step? Why do we equate the derivative to zero?

 

[hilbert2]

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 \frac{d}{dx}x^{n}=nx^{n-1} with n=-1, we get \frac{d}{dm}(-2/m) =-2\frac{d}{dm}m^{-1}=-2\times(-1)\times m^{-2}=2/m^{2}. 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.

 

[Ray Vickson]

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
\text{OP} + \text{OQ} = 10 + 8m + \frac{2}{m} \equiv S(m). 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
\frac{1}{2}(a+b) \geq \sqrt{a b} \Longrightarrow a+b \geq 2 \sqrt{ab}, with equality holding if and only if $a = b$. So, applying this we have
8m + \frac{2}{m} \geq 2 \sqrt{8m \frac{2}{m}} = 2 \sqrt{16} = 8, with equality holding only when $8m = 2/m,$ or $m^2 = 1/4$, or $m = \pm 1/2 \longrightarrow m = 1/2$.