Question

Eh.

Answer 1

Eh?

Answer 2

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

Answer 3

So I did this: I assumed that \(a,b > 0\) and the equation of L is \(L: \dfrac{x}{a} + \dfrac{y}{b} = 1\)
Another constraint is\[\frac{8}{a} + \frac{2}{b} = 1\]So I am asked to minimize \(a+b\) given \(a,b>0\) and \(8/a + 2/b =1\)

Answer 4

@rational how

Answer 5

I tried to minimize \((a+b)^2 \) as an intermediary.\[=a^2 + b^2 + 2ab = a^2 + b^2 + 2(8b + 2a) = a^2 + b^2 + 16b + 4a \]I'm not reachin anywhere.

Answer 6

Haha, maybe I am.

Answer 7

\[\dfrac{8}{a} = \dfrac{b-2}{b}\]\[a = \frac{8b}{b-2}\]So\[a+b = \dfrac{b^2 - 2b + 8b}{b-2}=\dfrac{b^2 + 6b}{b-2}\]

Answer 8

Got it.