find two positive numbers such thier product is 192 and sum of the first plus three times the second is a minimum

Question

i need help on this questions and i have no clue how to do it can yall help me?

Answer 1

Let x and y be the two positive numbers.

 

-  Their product is 192: x*y = 192

-  the sum of the first plus 3 times the second is a minimum:  x + 3y

 

From the first equation, y = 192/x.  Substitute that into the second equation:

 

x + 3y = x + 3(192/x) = x + 576/x

 

Now take the first derivative, set it to zero, and solve for x:

 

d(x + 576/x)/dx = 1 - 576/x²

0 = 1 - 576/x²

0 = x² - 576                         [Multiplied both sides by x²]

576 = x²

√576 = √x²

 

±24 = x

 

Since we want a positive number, x = 24.  y = 192/x = 8

 

As a check, the factors of 192 are 192*1, 96*2, 48*4, 24*8, 16*12

 

192 + 3*1 = 195

96 + 3*2 = 102

48 + 3*4 = 60

24 + 3*8 = 48

16 + 3*12 = 52

 

24 + 3*8 is indeed the minimum.