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

AB=192 if we devids 192 over 4 we'll get 48
a+3b=12+48=60
I think that's the minimum sum !! I don't think ther is a role on this kind of puzzle

Answer 2

A=8
B=24
3A+B=48

Answer 3

Start with what you know:

\[xy = 192\]\[z = x + 3y\]
We're interested in minimizing z, which changes as x and y change (a clue that you're going to be taking a derivative). Let's make it easier by focusing on one changing variable at a time. Solve the first equation for x (or y, but the math is a little easier for x) and plug into the second:

\[x = \frac{192}{y}\]\[z = \frac{192}{y} + 3y\]
Now take the derivative of both sides (think of 192/y as 192(y^(-1)) if that helps:

\[\frac{dz}{dy} = \frac{-192}{y^{2}} + 3\]
To find the minimum, set dz/dy to zero and solve:

\[0 = \frac{-192}{y^{2}} + 3\]\[-3 = \frac{-192}{y^{2}}\]\[-y^{2} = \frac{-192}{3}\]\[y^{2} = 64\]\[y = \pm 8\]
Since the original problem states positive numbers, the solution is y = 8. Now we can solve for x by plugging y back into the original equation:

\[8x = 192\]\[x = 24\]
So the solution is 8 and 24.