Find two positive numbers satisfying the given requirements. the product is 192 and the sum of the first plus three times the second is a minimum. (first number) (second number)

Two positive numbers x and y such that xy = 192

The values that minimize x + 3y

y = 192/x . . . . . solve for y

f(x) = x + 3y

f(x) = x + 3(192/x) . . . . . the function we want to minimize

We can find the x that minimizes of f(x) by setting the derivative of f(x) to zero.

... f'(x) = 1 - 576/x² = 0

... 576 = x² . . . . . . . . . . . . multiply by x², add 576

... √576 = x = 24 . . . . . . . take the square root

... y = 192/24 = 8 . . . . . . . find the value of y using the above equation for y

The first number is 24.

The second number is 8.

understand that i don't know the answers

will someone answer his question