Find three positive integers x, y, and z that satisfy the given conditions. The product is 125, and the sum is a minimum.

Step-by-step explanation:

Given 3 positive integers x, y and z, if their product is 125, then;

xyz = 125 1

Sum of the integers S(x, y, z) = x+y+z 2

From equation 1; x = 125/yz 3

Substitute equation 3 into 2

S(x, y, z) =  125/yz+y+z

Sy  =  -125/y²z + 1

Sz  =  -125/z²y + 1

For the sum to be minimum, Sy = Sz = 0

-125/y²z + 1 = 0

-125/y²z = -1

-125 = -y²z

125 = y²z

Also; is Sz = 0

-125/z²y + 1 = 0

-125/z²y = -1

-125 = -z²y

125 = z²y

Hence y²z = z²y = 125

From y²z = 125, z = 125/y²

Substitute z = 125/y² into the equation 125 = z²y

125 = (125/y²)²× y

125 = 125²/y⁴ × y

125 = 125²/y³

1 = 125/y³

y³ = 125.

y = 5

Since 125 = y²z

5²z = 125

25z = 125

z = 5

From the product xyz = 125

x(5)(5) = 125

25x = 125.

x = 5

Hence x = y = z = 5

i think the correct option is the second option.

12

step-by-step explanation:

12 + 5 = 17

12x2=24

24-7=17