1961 AMC 12 Problem 40

Question

Find the minimum value of

\sqrt{x^2+y^2}

if

5x+12y=60.

\dfrac{60}{13}

\dfrac{13}{5}

\dfrac{13}{12}

1

0

Interpret

\sqrt{x^2+y^2}

as distance from the origin

Apply Cauchy-Schwarz to

5x+12y

By Cauchy-Schwarz,

\begin{aligned} 60&=5x+12y\\ &\le\sqrt{5^2+12^2}\\ &\qquad\cdot\sqrt{x^2+y^2}\\ &=13\sqrt{x^2+y^2}. \end{aligned}

Thus the distance is at least

\frac{60}{13}.

Equality occurs when

(x,y)

is proportional to

(5,12),

so the minimum is

\frac{60}{13}.

Therefore, the correct answer is A.

Problem 40 in other years