Minimum Distances

Consider the image. From the properties of similar triangle we can say that $\frac{y}{3}= \frac{48}{x}\quad(How)$ $\Rightarrow xy=48×3$ $\Rightarrow OP•OQ=(3+x)(48+y)=48(x+\frac{9}{x}+6)$

Now from AM-GM, as $x$ is always positive, the minimum value of $(x+\frac{9}{x})$ is 6 (How).

Hence minimum value of $OP•OQ \space is \space 48(6+6)=\boxed{576}$

Let $OP=a$ and $OQ=b$ . By definition, the line that passes through $P=(a,0)$ and $Q=(0,b)$ is called the symmetric equation of the line, and it is: $\dfrac{x}{a}+\dfrac{y}{b}=1$ . We know that it passes through $(3,48)$ , hence: $\dfrac{3}{a}+\dfrac{48}{b}=1$

Applying AM-GM inequality we get:

$\dfrac{\dfrac{3}{a}+\dfrac{48}{b}}{2} \geq \sqrt{\dfrac{3}{a} \times \dfrac{48}{b}}$

$\dfrac{1}{2} \geq \sqrt{\dfrac{144}{ab}}$

$\dfrac{1}{4} \geq \dfrac{144}{ab}$

$ab \geq \boxed{576}$

$Let\quad A(3,48)\quad be\quad the\quad point\quad that\quad lies\quad on\quad the\quad reqd.\quad line.\\ \\ Now,\quad let\quad the\quad equation\quad of\quad the\quad line\quad be\quad y\quad =\quad mx+c,\\ where\quad 'm'\quad is\quad the\quad slope\quad of\quad the\quad line,\quad and\quad 'c'\quad is\quad the\\ Y-intercept(OQ).\\ Slope\quad of\quad a\quad line\quad is\quad given\quad by,\quad \frac { y-{ y }_{ 1 } }{ x-{ x }_{ 1 } } ,\quad where\quad ({ x }_{ 1 }{ ,y }_{ 1 })\quad is\\ any\quad point\quad on\quad the\quad line.\\ \\ In\quad our\quad case,\quad m\quad =\quad \frac { y-48 }{ x-3 } .\\ Using\quad the\quad value,\quad y\quad =\quad (\frac { y-48 }{ x-3 } )x+c\\ xy-3y\quad =\quad \quad xy-48x+cx-3c\\ Arranging\quad the\quad terms,\\ 48x-3y\quad =\quad cx-3c.\\ \\ Now\quad to\quad find\quad the\quad x-intercept(OP),\quad put\quad y=0\\ 48x-0\quad =\quad cx-3c\\ x\quad =\quad \frac { 3c }{ c-48 } \\ OP\times OQ\quad =\quad \frac { 3{ c }^{ 2 } }{ c-48 } \\ To\quad find\quad the\quad minimum\quad of\quad OP\times OQ,\\ it's\quad derivative\quad =\quad 0\\ \\ \frac { d }{ dc } (OP\times OQ)\quad =\quad 0\\ \\ \frac { d }{ dc } \frac { 3{ c }^{ 2 } }{ c-48 } \quad =\quad 0\\ \\ \frac { d }{ dc } { c }^{ 2 }({ c-48) }^{ -1 }\\ \\ { c }^{ 2 }\frac { d }{ dc } { (c-48) }^{ -1 }+{ (c-48) }^{ -1 }\frac { d }{ dc } { c }^{ 2 }\quad =\quad 0\\ \frac { -{ c }^{ 2 } }{ { (c-48) }^{ 2 } } +\frac { 2c }{ c-48 } \quad =\quad 0\\ Solving\quad this\quad equation\quad 'c'\quad comes\quad out\quad to\quad be\quad 0\quad or\quad 96\\ Since,\quad it\quad is\quad given\quad that\quad it\quad cuts\quad positive\quad intercepts,\\ c=0\quad is\quad rejected.\\ \\ So\quad c=96,\quad will\quad give\quad the\quad minimum\quad value\quad of\quad OP\times OQ\\ { OP\times OQ }_{ min. }\quad =\quad \frac { 3\quad \times \quad 96\quad \times \quad 96 }{ 96-48 } \\ \quad =\quad 576$

CHEERS!:):)

Suppose the line takes the form $ax + by = c.$ Then we know $c = 3a + 48b.$ Solving for our intercepts we get

(0, $\frac{3a}{b} + 48$ ) and (3 + $\frac{48b}{a}$ , 0).

Thus OP * OQ = $(\frac{3a}{b} + 48) * (\frac{48b}{a} + 3) \ge (2 * \sqrt{48*3})^2 = 576$ by Cauchy-Schwarz.