Changing It Up

Consider the image. From the properties of similar triangle we can say that $\frac{y}{3}= \frac{48}{x}\quad(How)$ $\Rightarrow xy=48×3$ $\Rightarrow OP+OQ=(3+x)+(48+y)=51+x+\frac{144}{x}$

Now from AM-GM, as $x$ is always positive, the minimum value of $(x+\frac{144}{x})$ is 24 (How).

Hence minimum value of $OP+OQ \space is \space 51+(24)=\boxed{75}$

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 Titu's Lemma we get:

$\dfrac{3}{a}+\dfrac{48}{b} \geq \dfrac{(\sqrt{3}+\sqrt{48})^2}{a+b}$

$1 \geq \dfrac{(5\sqrt{3})^2}{a+b}$

$a+b \geq \boxed{75}$

$Let~~~y=m(x-x_o)+y_o ~~be~the~line,~~(x_o,y_o)=(3,48).\\ \therefore~~~y~-~m*x~=~-~m*x_o~+~y_o.~~~~~~~~~\implies~~\dfrac y {-~m*x_o~+~y_o}~-~\dfrac x {-~m*x_o~+~y_o}=1.\\ Equation~of~a~line~is~~\dfrac x a~+~\dfrac y b=1. \\ So~the~sum ~of ~intercepts~~f(m)=~y-intercpet + x-intercpet ~=~a+b~= ~(-~m*x_o~+~y_o)~-~(-~x_o~+~\dfrac{y_o} m). \\ f(m)_{(3,48)}=~(-3m+48)~~-~~(-3+\dfrac{48} m).\\ So~f ' (m)=~-3~- \dfrac {48}{m^2}=0. ~~~~m^2=16.~~\implies~~m=-4,~~in~ 1st~quadrant.\\ \therefore~f(-4)=~(4*3+48)~~-(-3~-~\dfrac{48} {-4}) ~=~\Large \color{#D61F06}{75}.$

First, Some definitions. I'm going to let $p$ be the distance of $OP$ and $q$ be the distance of $OQ$ . Similarly, I will let $T$ be the value of $p+q$ . The question asks to find the smallest value of $OP + OQ$ which in other words will mean finding the smallest value of $T$

Let's write the line as an equation going through point $(3,48)$ . We don't know the gradient of the line so let's call it $m$ . Now using the form $y - y_{1} = m(x-x_{1})$ , the line can be written as $y-48 = m(x-3)$ .

As $P$ and $Q$ represent the points where the line crosses the axes, we can also represent them as $(p, 0)$ and $(0, q)$ respectively. By substituting these points into the line equation, we obtain $0-48 = m(p-3) \Rightarrow pm = 3m-48$ and $q-48 = m(0-3) \Rightarrow qm = -3m^2 + 48m$ Adding these two equations together gives $pm + qm = Tm = -3m^2 + 51m - 48 \Rightarrow T = -3m + 51 - \frac{48}{m}$

(accidently clicked submit. finishing it off now.)

the line equation

y = m x + c ---> 48 = 3 m + c ----> c = 48 - 3 m

the length of "op + oq" = -c /m + c = - (48 - 3 m)/m + (48 - 3 m)

Find the minimum by differentiating with respect to "m" = 0

this leads to -3 + 48 / m^2 = 0 -- slove ----> m = -4 (min)

the line equation becomes y = m x +c = -4 x + 60

op + oq = 15 + 60 = 75