Answer to Question #142448 in Calculus for Peace College

Question

120 numbers are written on the board. It is known that among all their pairwise
products there are exactly 2000 negative numbers. What is the largest number of
zeroes that could be written on the board?

Answer 1

Let $x=$ the number of negative numbers, $y=$ the number of positive numbers, and $z=$ the number of zeros. Then

$$x+y+z=120$$

It is known that among all their pairwise products there are exactly 2000 negative numbers.

$$xy=2000$$

$$y=\dfrac{2000}{x}$$

Hence

$$z=z(x)=120-x-\dfrac{2000}{x}$$

Find the first derivative with respect to $x$

$$z'(x)=(120-x-\dfrac{2000}{x})'=-1+\dfrac{2000}{x^2}$$

Find the critical number(s):

$$z'(x)=0=>-1+\dfrac{2000}{x^2}=0$$

$$x^2=2000$$

$$x_1=-20\sqrt{5}, x_2=20\sqrt{5}$$

Critical numbers: $-20\sqrt{5},20\sqrt{5}.$

We consider $0<x<120$

First Derivative Test

If $0<x<20\sqrt{5},$ then $z'(x)>0, z(x)$ increases.

If $20\sqrt{5}<x<120,$ then $z'(x)<0,z(x)$ decreases.

The function $z(x)$ has a local maximum at $x=20\sqrt{5}.$

Since the function $f(x)$ has the only extremum on $(0, 120),$ then the function $z(x)$ has the absolute maximum on $(0,120)$ at $x=20\sqrt{5}.$

$$20\sqrt{5}\approx44.72$$

$$44<44.72<45$$

The factors of 2000 nearest to 44.72 are $40$ and 50

$$z(40)=120-40-\dfrac{2000}{40}=30$$

$$z(50)=120-50-\dfrac{2000}{50}=30$$

Therefore he largest number of zeroes that could be written on the board is 30.