The Easy Minimum

Before reading further it would be good to have a look at this wiki .

---

$\displaystyle \sqrt { { x }^{ 2 }+{ y }^{ 2 } }$ represents the distance between Origin $\displaystyle \left( 0,0 \right)$ and the point $\displaystyle\left( x,y \right)$ . This point, i.e $\displaystyle\left( x,y \right)$ , also lies on the line $\displaystyle 5x+12y=60$ .Now the minimum distance of a point from a line is the perpendicular distance from it. Hence, $\displaystyle \sqrt { { x }^{ 2 }+{ y }^{ 2 } }$ represents the perpendicular distance of Origin $\displaystyle \left( 0,0 \right)$ from the line $\displaystyle 5x+12y=60$ . Which is equal to $\displaystyle \cfrac { \left| 60-5\times 0-12\times 0 \right| }{ \sqrt { { 5 }^{ 2 }+{ 12 }^{ 2 } } } =\cfrac { 60 }{ 13 }$ . Hence the answer.

Another approach is to use Cauchy-Schwarz inequality:

$(5^2+12^2)(x^2+y^2) \geq (5x+12y)^2 \\ 13^2(x^2+y^2) \geq 60^2 \\ x^2+y^2 \geq \dfrac{60^2}{13^2} \\ \sqrt{x^2+y^2} \geq \boxed{\dfrac{60}{13}}$

The equation can be rewritten as $\frac{x}{12}+\frac{y}{5}=1$ , which represents a line with $x$ -intercept $12$ and $y$ -intercept $5$ . So the line and the axes form a $5-12-13$ right triangle. The minimum value $h$ of $\sqrt{x^2+y^2}$ is the height of the triangle with base $13$ . Equating twice the area of the triangle in two different ways,

$13h = 5\times12$ ,

$h = \frac{60}{13}$ .

---

[Latex edits - moderator]

I recently learned Lagrange multipliers in my multivariable class. I decided to have fun with it. Lagrange multipliers find the relative maxes and mins based on a constraint on an expression.

The constraint is $5x + 12y = 60$ , and we want to minimize $\sqrt{x^2+y^2}$

The partial derivatives of the expression and the constraint gives these two equations, where $\lambda$ is the multiplier:

$\frac {1}{2 \times \sqrt{x^2+y^2}} \times 2x = \lambda \times 5$

$\frac {1}{2 \times \sqrt{x^2+y^2}} \times 2y = \lambda \times 12$

$\frac {1}{2 \times \sqrt{x^2+y^2}}$ can be referred to as f(x,y). Thus: $12 \times f(x,y) \times 2x = 5 \times f(x,y) \times 2y$

f(x,y) cannot equal zero, so the equation simplifies to: $12x = 5y$ Plugging this back in, we can solve for either y or x. Substituting for x, $\frac{25y}{12}+12y=60$ .

$y=\frac{720}{169}$ and $x=\frac{300}{169}$ .

When plugged into the expression $\sqrt{x^2+y^2}$ , the answer comes out to be 60/13.

To be honest I have no idea where the max is. Thank god for multiple choice questions.

EDIT: After reading other solutions, it seems pretty easy now that it is the distance from the line. Creativity takes effort, and I don't like effort when I wake up. Mindlessly calculating is much easier. In this case, there wouldn't be a max because the maximum value is infinity.

another solution can be putting x=a which gives y= (60-5a)/12. plugging this in other equation which will be a quadratic in 'a' and the min(quadratic) is a well known exercise for mathsies.. !! :D