what are three numbers whose product is 100 and the sum of their squares is less than 65?

Question

the three numbers are positive.

Answer 1

I thought you could do it with integers:
4 * 5 * 5 = 100
4² + 5² + 5² = 66

But that didn't work out. So that means you have to use numbers that aren't integers. The cube root of 100 (4.64158883) would seem like a good choice. That would minimize the sum of the squares...

4.6416 * 4.6416 * 4.6416 = 100
4.6416² + 4.6416² + 4.6416² ≈ 64.63 < 65

This is the minimum answer, but you could actually allow for some variation in the numbers and still be under 65.

For example:
4.6 * 4.6 * 4.725898 = 100
4.6² + 4.6² * 4.725898² ≈ 64.65 < 65

4.5 * 4.5 * 4.9382 = 100
4.5² + 4.5² + 4.9382² ≈ 64.89 < 65

4.8 * 4.6 * 4.528986 = 100
4.8² + 4.6² + 4.528986² ≈ 64.71 < 65

But generally, all your answers would need to include numbers as close as possible to cuberoot (100).

Answer 2

All three numbers are the cube root of 100. By definition, (100^(1/3))*(100^(1/3))*(100^(1/3)) = 100. And by calculation, (100^(1/3))^2 + (100^(1/3))^2 + (100^(1/3))^2 = 64.633, which is less than 65.

In general, something like the sum of numbers with a certain product (or the sum of their squares) is minimized when the numbers are as close to each other as possible. Note that 1*100, 2*50, 4*25, 5*20, and 10*10 all have products of 100, but the sums of the pairs of numbers are 101, 52, 29, 25, and 20, descending as the two numbers approach each other. So for a rough cut at this problem, you could try 4*5*5 = 100. However, the sum of the squares is 16 + 25 + 25 = 66, too large. There are no triples of integers that have a product of 100 that are closer in value, so it became clear that I had to abandon integers and take the cube root directly, giving three factors that are equal, guaranteeing that I would minimize the square sum.

Answer 3

Maximize f(x,y,z)=xyz subject to x+y+z=one hundred g(x,y,z)=x+y+z-one hundred Lambda(x,y,z,lambda) = f(x,y,z) + lambda g(x, y, z) = xyz + lambda (x+y+z-one hundred) locate the derivatives with admire to x,y,z and lambda (regardless of the shown fact that it somewhat is the unique constraint) as quickly as you remedy the derivatives, replace to locate the bright factors. Whichever extreme factors yield maximums are the respond.

Answer 4

xyz=100
x^2+y^2+z^2<65

x=y=z=100^(1/3)=4.6415
check
4.6415^2=21.544
3*21.544=64.633