Question

can someone explain how to find two positive numbers such that their product is 192 and the sum of the first plus three times the second is a minimum?

Answer 1

Let the two numbers be y and x, then
xy = 192
y = 192/x
dy/dx = -192/x^-2 . . . (1)

Sum (S) = y + 3x
If the sum is minimum, then
dS/dx = dy/dx + 3 = 0
dy/dx = -3

From (1), -192/x^2 = -3
3x^2 = 192
x^2 = 192/3 = 64
x = sqrt(64) = 8
y = 192/8 = 24

The two positive numbers are 8 and 24.