two positive numbers so that the product is 192 and the some of the first plus three times the second is min?

Answer 1

xy= 192 z= x+3y min.
z= x+576/x ´
z´= 1 -576/x^2 =0 ===> x^2 = 576 and x= 24 y=8

z´= (x2-576)/x2 and the sign is + -24 - 24 + so it´s a minimum

Answer 2

enable x be one among the two numbers. this implies the different is 192/x. The sum of the two numbers, f(x) is f(x) = x + 192/x it relatively is minimum takes place the place f '(x) = 0 and f ''(x) > 0 f '(x) = a million - 192/x^2 0 = a million - 192/x^2 192/x^2 = a million 192 = x^2 x = ?192 or x = -?192 The question requests 2 valuable numbers so discard x = -?192, leaving purely x = ?192 = 8?3 f ''(x) = 384/x^3 f ''(8?3) = 384/(8?3)^3 > 0 so it relatively is a minimum the different quantity is 192/(8?3) = 8?3. the two valuable numbers are 8?3 and eight?3. 2nd situation: the two numbers are x and 192/x Sum = f(x) = x + 3(192/x) f '(x) = a million - 576/x^2 0 = a million - 576/x^2 x^2 = 576 x = 24 the different quantity is 192/24 = 8

Answer 3

12 and 16

Answer 4

xy =192
Sum = S = x+3y
S = x + 3(192/x) = x+ 576/x
dS/dx = 1 - 576/x^2
0 =1- 576/x^2
-1= -576/x^2
-x^2= -576
x = 24
y = 8

Answer 5

24 and 8
24 * 8 = 192
24 + 3(8) = 48

Answer 6

Call them X & Y.
XY=192
X+3Y=min ???

min?