Question

We have a straight line passing through the point \((h,k)\) in the first quadrant. It intersects the positive x and y axes at points A and B. Find the minimum value of:

i. OA + OB
ii. OA x OB

Answer 1

O is the origin.

Answer 2

\[\dfrac{x}{a} + \dfrac{y }{b} = 1\]\[\dfrac{h}{a} + \dfrac{k}{b} = 1 \tag{a, b > 0}\]\[\Rightarrow hb + ka = ab\]\[\Rightarrow hb = ab - ka\]\[\Rightarrow b = \dfrac{ab - ak}{h}\]

Answer 3

Part 1:\[\min (OA + OB) = \min (a + b)\]\[a + b = a + \dfrac{ab - ak}{h} = a\left(1 + \frac{b - k}{h`}\right)\]

Answer 4

Eh...

Answer 5

Maybe solving for the general case doesn't help. The original question says \((h,k) \equiv (3,48)\).

Answer 6

\[\dfrac{3}{a} + \dfrac{48}{b} = 1\]\[\dfrac{48}{b} = \dfrac{a - 3}{a}\]\[b = \dfrac{48a}{a-3}\]

Answer 7

There.

Answer 8

\[a+b = a + \dfrac{48a }{a-3}\]\[= \dfrac{a^2 - 3a + 48a}{a-3}\]\[=\dfrac{a^2 + 45a}{a-3}\]

Answer 9

How do I minimize this function given \(a > 0\)?

Answer 10

\[b> 0 \Rightarrow 48a(a-3) > 0 \Rightarrow a(a-3)>0 \Rightarrow a \in (-\infty , 0) \cup (3 , \infty)\]

Answer 11

So taking the intersection,\[a\in (3,\infty)\]

Answer 12

h,k are constants, a,b are the variables.